Exploring Algebraic Identities

Pick any three consecutive square numbers — say $1, 4, 9$ — add the smallest and the largest, then subtract twice the middle one: $1 + 9 - 2(4) = 2$ . Try $25, 36, 49$ : $25 + 49 - 2(36) = 74 - 72 = 2$ . Try $100, 121, 144$ and the answer is still $2$ . That is not a coincidence waiting to be explained away — it is a hidden algebraic identity waving at you, and this chapter builds the machinery that turns such tricks from magic into two lines of proof. An algebraic identity is an equation that is true for every value of the variables in it, unlike an ordinary equation, which is true only for particular values. $x^2 - 1 = 24$ is an equation, true only when $x = 5$ or $x = -5$ ; but $(x+y)^2 = x^2 + 2xy + y^2$ is an identity, true for every $x$ and $y$ you can name. This chapter collects the identities you met in Class 8 — the square of a sum and a difference — and pushes them further, first geometrically, as areas cut and rearranged, then algebraically, by repeated use of the distributive property, so no picture ever has to be trusted blindly. From there the chapter climbs two ways. Sideways, the square-of-a-sum identity grows to three terms, $(a+b+c)^2$ , and a companion identity attributed to Śhrīdharāchārya, $a^2 = (a+b)(a-b)+b^2$ , turns squaring numbers like $55$ into fast mental arithmetic. Upward, the same ideas extend one dimension higher, to the cube of a sum and difference, visualised as a cube sliced into smaller cubes and cuboids, then to the sum and difference of cubes, and finally to a three-variable identity for $x^3+y^3+z^3-3xyz$ . Every identity does double duty: it expands products instantly, and read backwards it factors expressions — by pattern-matching, then by splitting the middle term, made concrete with algebra tiles. The chapter closes by using this toolkit to simplify rational expressions and solve real problems. By the end, the always-2 trick above will look obvious, because you will be able to prove it in your head.

Key takeaways

An algebraic identity is an equation true for every value of its variables; an equation such as $x^2-1=24$ is true only for specific values.
Square of a sum/difference: $(a\pm b)^2 = a^2\pm2ab+b^2$ — provable by the distributive law and visualised as a square cut into four pieces.
Square of a trinomial: $(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca$ , found by treating $b+c$ as one block and reusing the square-of-a-sum identity.
Difference of squares: $a^2-b^2=(a+b)(a-b)$ , rearranged as $a^2=(a+b)(a-b)+b^2$ to square numbers near a round base mentally, e.g. $55^2 = 60\times50+25=3025$ .
Quadratics $x^2+(a+b)x+ab$ factor as $(x+a)(x+b)$ by splitting the middle term — choosing $a,b$ so their sum matches the middle coefficient and product matches the constant.
Algebra tiles turn that factoring into a picture: an $x^2$ -tile, $x$ -tiles and unit tiles rearrange into a rectangle whose sides are the two linear factors.
Cube identities: $(a\pm b)^3=a^3\pm3a^2b+3ab^2\pm b^3$ , visualised as a cube of edge $(a+b)$ split into two cubes and six cuboids.
Sum/difference of cubes: $x^3\pm y^3=(x\pm y)(x^2\mp xy+y^2)$ , found by multiplying out and checking terms cancel.
The three-variable identity $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ finds a sum of cubes from a sum, a product and a sum of squares, without cubing any number directly.
Key idea: every identity in this chapter is the distributive law applied once and for all to a fixed pattern — (a plus b), (a plus b plus c), or a cube — so recognising which pattern an expression matches instantly hands you its expansion, its factors, or a fast mental calculation.

The core insight

Algebraic identities such as $(a\pm b)^2$ , $a^2-b^2$ , $(a\pm b)^3$ and $x^3+y^3+z^3-3xyz$ are equations true for every value of their variables, each derivable once from the distributive property, and each reusable to expand expressions, factorise them by matching or by splitting the middle term, compute squares, cubes and products mentally, and simplify rational algebraic expressions by cancelling common factors.

What this chapter covers

The chapter builds outward from one question and one tool. The question, raised by the always-2 pattern of consecutive squares, is what makes a statement true for every number rather than just a few — the identity-versus-equation distinction the chapter rests on. The tool is the distributive property, reused to prove each new identity from the ones before it.

4.1–4.2 build $(a+b)^2$ and $(a-b)^2$ through the geometry of squares and rectangles, proved by area and by algebra, and used to square numbers mentally.
4.3 turns those same identities backward, to factorise quadratics that are perfect squares.
4.4 extends the square identity to three terms, $(a+b+c)^2$ , and revisits the difference-of-squares identity as a fast squaring trick.
4.5–4.6 cover the general method for factorising $x^2+px+q$ by splitting the middle term, first with algebra tiles, then purely algebraically.
4.7 finds identities for cubes: $(a\pm b)^3$ , the sum and difference of cubes, and the three-variable identity for $x^3+y^3+z^3-3xyz$ .
4.8 simplifies rational algebraic expressions by cancelling common factors.

Worked examples below reuse NCERT's own example numbers where named, so you can cross-check against your textbook.

The core ideas

Identity versus equation

Every identity in this chapter is a special kind of equation — one where both sides describe the same quantity, written two different ways, so equality holds no matter what you substitute. An ordinary equation like $x^2-1=24$ pins down specific values of $x$ (here $x=5$ or $x=-5$ ); an identity like $(x+y)^2=x^2+2xy+y^2$ holds for every pair $x,y$ , including negatives and fractions. Check it: with $a=-2,b=-3$ , the left side $(a+b)^2=25$ and the right side $a^2+2ab+b^2=4+12+9=25$ agree. But checking cases only builds confidence — only an argument using the distributive property, valid for general $a,b$ , actually proves an identity.

The square of a sum

Take two line segments of lengths $a$ and $b$ and build a square of side $(a+b)$ . Splitting it by a horizontal and a vertical line at distance $a$ from one corner divides it into a square of side $a$ , a square of side $b$ , and two rectangles of sides $a$ and $b$ . Adding the four areas gives the identity.

(a+b)^2 = a^2 + 2ab + b^2.

where: $a$ and $b$ are any two quantities, and the middle term $2ab$ comes from the two equal rectangles of area $ab$ in the picture. This is confirmed algebraically: $(a+b)^2 = (a+b)(a+b) = a(a+b)+b(a+b) = a^2+ab+ba+b^2 = a^2+2ab+b^2$ , an argument that works for every $a$ and $b$ , not only lengths. The identity expands binomials, for instance $(5x+2y)^2 = 25x^2+20xy+4y^2$ , and speeds up mental squaring: $43^2 = (40+3)^2 = 1600+240+9 = 1849$ .

Figure

A large square of side a plus b split into a square of side a, a square of side b, and two equal rectangles of sides a and b, illustrating that the total area equals a squared plus 2ab plus b squared

The square of a difference

Split a square of side $a$ into a smaller square of side $(a-b)$ and two rectangles, by marking a length $b$ off one side. Subtracting the two rectangle areas from the big square gives a companion identity.

(a-b)^2 = a^2 - 2ab + b^2.

where: $a$ is the side of the larger square and $b$ the piece removed, so $(a-b)$ is what remains; the middle term is subtracted because the two rectangles were double-counted once the small square's area was already included. Algebraically, replacing $b$ by $-b$ in the square-of-a-sum identity gives this result directly, and it explains the overview's consecutive-squares trick: $(n-1)^2+(n+1)^2=2n^2+2$ , so subtracting $2n^2$ always leaves $2$ . It is equally useful for mental arithmetic: $29^2=(30-1)^2=900-60+1=841$ .

The square of a trinomial

To square a sum of three terms, treat two of them as a single block and reuse the square-of-a-sum identity twice. Writing $d = b+c$ turns $(a+b+c)^2$ into $(a+d)^2 = a^2+2ad+d^2$ , and expanding $d^2=(b+c)^2$ separately finishes the job — the full derivation is worked through step by step below.

(a+b+c)^2 = a^2+b^2+c^2+2ab+2bc+2ca.

where: the three squared terms $a^2, b^2, c^2$ and the three cross terms $2ab, 2bc, 2ca$ each appear once, so a trinomial squared has six terms before any combine. This squares three-digit numbers by splitting into hundreds, tens and units: $119^2 = (100+10+9)^2 = 10000+100+81+2000+1800+180 = 14161$ .

The difference of squares

A sum multiplied by the difference of the same two terms collapses neatly, because the two middle terms of the expansion are equal and opposite and cancel.

a^2 - b^2 = (a+b)(a-b).

where: $a$ and $b$ are the two terms, and the identity works only when both brackets share the same $a$ and $b$ with opposite signs — $(2x+3)(2x-5)$ does not match, since the second terms differ. Rearranged as $a^2 = (a+b)(a-b)+b^2$ , this identity — recorded by Śhrīdharāchārya around 750 CE — squares a number near a round base fast: $55^2 = (55+5)(55-5)+5^2 = 60\times50+25 = 3025$ , using $b=5$ to reach the round product $60\times50$ .

Factorising quadratics that are perfect squares

Reading the square identities right to left turns them into factoring tools: a trinomial matching $a^2\pm2ab+b^2$ factors instantly as a single squared bracket.

x^2+4x+4 = x^2+2(x)(2)+2^2 = (x+2)^2, \qquad 36x^2+12x+1 = (6x)^2+2(6x)(1)+1^2 = (6x+1)^2.

where: matching means identifying which term is $a^2$ , which is $b^2$ , and checking the middle term is exactly $2ab$ ; if a common numerical factor hides the pattern, as in $50p^2+60pq+18q^2 = 2(25p^2+30pq+9q^2)$ , remove it first, then match: $2(5p+3q)^2$ .

Factorising quadratics by splitting the middle term

Not every quadratic is a perfect square, so a general method is needed for $x^2+px+q$ . Write the middle coefficient $p$ as a sum $a+b$ and the constant $q$ as the matching product $ab$ , then group the four resulting terms in pairs.

x^2+(a+b)x+ab = (x+a)(x+b).

where: $a$ and $b$ must satisfy both $a+b=p$ and $ab=q$ at once — matching only one condition is not enough. For $x^2+11x+30$ , the factor pairs of $30$ are checked against a sum of $11$ until $a=5, b=6$ works, giving $(x+5)(x+6)$ . When $p$ is negative, as in $x^2-5x+6$ , both $a$ and $b$ must be negative, giving $a=-2, b=-3$ .

Factorising with algebra tiles

The splitting trick has a physical picture. Represent $x^2+7x+12$ using one $x^2$ -tile, seven $x$ -tiles and twelve unit tiles, then rearrange them into a single rectangle. The $7x$ term must split as $3x+4x$ so three $x$ -tiles sit along one side of the $x^2$ -tile and four along the adjacent side, with the twelve unit tiles filling a $3$ -by- $4$ array in the corner — only then do the tiles close into a rectangle whose sides $(x+3)$ and $(x+4)$ are the two factors.

(x+3)(x+4) = x^2+3x+4x+12 = x^2+7x+12.

where: the split of $7x$ into $3x+4x$ is not arbitrary — only the pair $3$ and $4$ , product $12$ and sum $7$ , lets the tiles form a rectangle; any other split, such as $2x+5x$ , leaves gaps and cannot be tiled cleanly.

Figure

A rectangle built from algebra tiles representing x squared plus 7x plus 12, showing one large x squared tile at top left, three x tiles along the top, four x tiles down the left side, and twelve unit tiles arranged in a 3 by 4 grid, with the whole rectangle having side lengths x plus 3 and x plus 4

The cube of a sum and a difference

Multiplying $(a+b)^2$ by one more factor of $(a+b)$ extends the pattern one dimension up. Geometrically, a cube of edge $(a+b)$ splits into two smaller cubes of volumes $a^3$ and $b^3$ , plus six cuboids: three of dimensions $a\times a\times b$ and three of $a\times b\times b$ .

(a+b)^3 = a^3+3a^2b+3ab^2+b^3, \qquad (a-b)^3 = a^3-3a^2b+3ab^2-b^3.

where: the coefficients $1,3,3,1$ match the piece count in the dissection — one $a^3$ -cube, three $a^2b$ -cuboids, three $ab^2$ -cuboids, one $b^3$ -cube. The second identity follows from the first by replacing $b$ with $-b$ ; the four terms then alternate in sign because odd powers of $-b$ stay negative while even powers turn positive.

Sum and difference of cubes

Multiplying out $(x-y)(x^2+xy+y^2)$ , every cross term except two cancels, leaving only a difference of cubes; the companion product $(x+y)(x^2-xy+y^2)$ behaves the same way and leaves a sum of cubes.

x^3-y^3=(x-y)(x^2+xy+y^2), \qquad x^3+y^3=(x+y)(x^2-xy+y^2).

where: the second factor is not a perfect-square trinomial — its middle term is $xy$ , not $2xy$ — so it cannot simplify further. These identities also explain why $x-y$ always divides $x^n-y^n$ for every whole number $n$ .

The three-variable identity for sums of cubes

Multiplying $(x+y+z)$ by $(x^2+y^2+z^2-xy-xz-yz)$ term by term produces eighteen terms that collapse dramatically, since twelve of them cancel in pairs, leaving only the three cubes and $-3xyz$ .

x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx).

where: a useful special case: whenever $x+y+z=0$ , the right side vanishes, so $x^3+y^3+z^3=3xyz$ automatically. Combined with the trinomial-square identity for $(x+y+z)^2$ , this finds a sum of cubes purely from a sum, a product and a sum of squares, as in the HOTS example below.

Simplifying rational algebraic expressions

A rational algebraic expression is a fraction whose numerator and denominator are polynomials; it simplifies the way a numerical fraction does, by factorising top and bottom and cancelling any factor common to both, provided that factor is never zero.

\frac{x^2-7x+12}{5x^2+5x-100} = \frac{(x-3)(x-4)}{5(x-4)(x+5)} = \frac{x-3}{5(x+5)}, \quad 5x^2+5x-100 \ne 0.

where: the numerator factors by splitting the middle term ( $a+b=-7$ , $ab=12$ gives $a=-3,b=-4$ ); the denominator first has $5$ removed, then splits again ( $a+b=1$ , $ab=-20$ gives $a=5,b=-4$ ). Only the entire factor $(x-4)$ , common to both, may be cancelled — individual terms inside a sum never can.

Deriving the identities

Derivation ADeriving the square of a trinomial

The goal is to find and prove a formula for $(a+b+c)^2$ , starting from nothing but the already-proved two-term identity $(a+d)^2=a^2+2ad+d^2$ . The strategy is common in algebra: reduce an unfamiliar three-term problem to a familiar two-term one by hiding two of the three terms inside a single new letter, solving that, then revealing what the letter was made of.

Step 1. Introduce a new letter $d$ standing for the sum of the last two terms: let $d=b+c$ . The expression to square, $a+b+c$ , is now exactly $a+d$ , a plain two-term sum whose square is already known.

Step 2. Apply the known square-of-a-sum identity to $a+d$ : $(a+d)^2=a^2+2ad+d^2$ , the same identity proved earlier for two terms, applied here to $a$ and $d$ .

Step 3. Substitute $d=b+c$ back into every occurrence of $d$ . The middle term becomes $2ad=2a(b+c)=2ab+2ac$ , and the last term $d^2=(b+c)^2$ needs expanding by the same identity: $(b+c)^2=b^2+2bc+c^2$ .

Step 4. Assemble the pieces: $(a+b+c)^2 = a^2+(2ab+2ac)+(b^2+2bc+c^2) = a^2+b^2+c^2+2ab+2bc+2ca$ . Every term is accounted for exactly once, so the identity holds for every choice of $a,b,c$ — not merely the geometric case, since the proof used only algebra, never a picture.

Step 5. Sanity-check with $a=1, b=2, c=3$ : the left side is $(1+2+3)^2=36$ , and the right side is $1+4+9+2(2)+2(6)+2(3)=36$ . Both agree, confirming Steps 1 to 4.

Go deeper · Derivation B · 1 minDeriving the cube of a sum and the cube of a difference

Derivation BDeriving the cube of a sum and the cube of a difference

The goal is to extend the square identities one power higher, proving $(a+b)^3$ and $(a-b)^3$ using only the distributive property and the square-of-a-sum identity, then checking the sign pattern in $(a-b)^3$ makes sense rather than being memorised blindly.

Step 1. Write $(a+b)^3$ as $(a+b)^2$ times one more factor of $(a+b)$ : $(a+b)^3=(a^2+2ab+b^2)(a+b)$ , using the square-of-a-sum identity already proved.

Step 2. Distribute $(a+b)$ across every term of $a^2+2ab+b^2$ : $a^2(a+b)=a^3+a^2b$ , then $2ab(a+b)=2a^2b+2ab^2$ , then $b^2(a+b)=ab^2+b^3$ .

Step 3. Add and collect like terms: $a^3+a^2b+2a^2b+2ab^2+ab^2+b^3 = a^3+3a^2b+3ab^2+b^3$ , the identity for $(a+b)^3$ .

Step 4. To find $(a-b)^3$ , replace every $b$ in the Step 3 result with $(-b)$ , since $a-b=a+(-b)$ : $[a+(-b)]^3 = a^3+3a^2(-b)+3a(-b)^2+(-b)^3$ .

Step 5. Resolve the sign of each power of $(-b)$ : $(-b)^2=+b^2$ (even power of a negative is positive), $(-b)^3=-b^3$ (odd power stays negative). This gives $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ — the same four terms as $(a+b)^3$ , alternating in sign because odd powers of $b$ flip sign while even powers do not.

Step 6. Verify with $a=5,b=1$ : $(5+1)^3=216$ matches $125+75+15+1$ , confirming Steps 1–3; and $(5-1)^3=64$ matches $125-75+15-1$ , confirming Steps 4–5.

Worked examples

Board level - Expanding and squaring with the square-of-a-sum identity

Expand $(5x+2y)^2$ , and use the same identity to evaluate $68^2$ without long multiplication.

Step 1. Match $(5x+2y)^2$ against $(a+b)^2=a^2+2ab+b^2$ with $a=5x, b=2y$ : $(5x)^2+2(5x)(2y)+(2y)^2 = 25x^2+20xy+4y^2$ .

Step 2. For $68^2$ , write $68=70-2$ , matching the square-of-a-difference pattern with $a=70, b=2$ .

Step 3. Apply the identity: $68^2=(70-2)^2=4900-280+4$ . $(5x+2y)^2=25x^2+20xy+4y^2$ and $68^2=\mathbf{4624}$ .

Board level - Factorising a quadratic by splitting the middle term

Factorise $x^2+11x+30$ (this reuses NCERT's own Example 11).

Step 1. Compare with $x^2+(a+b)x+ab$ : need $a+b=11$ and $ab=30$ .

Step 2. Check factor pairs of $30$ : $2\times15$ sums to $17$ , $3\times10$ sums to $13$ , but $5\times6$ sums to exactly $11$ .

Step 3. Split and factor by grouping: $x^2+5x+6x+30 = x(x+5)+6(x+5) = (x+5)(x+6)$ . $x^2+11x+30 = \mathbf{(x+5)(x+6)}$ .

Higher-order thinking (HOTS / olympiad-style) - Sum of cubes from a sum, a product and a sum of squares

The sum of three numbers is $10$ , their product is $25$ , and the sum of their squares is $38$ . Find the sum of their cubes without finding the numbers themselves (this reuses NCERT's own Example 15).

Step 1. Name the numbers $x,y,z$ : $x+y+z=10$ , $xyz=25$ , $x^2+y^2+z^2=38$ . The target $x^3+y^3+z^3$ needs the three-variable identity, which also needs $xy+yz+zx$ — not given directly.

Step 2. Find $xy+yz+zx$ from $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)$ : $100=38+2(xy+yz+zx)$ , so $xy+yz+zx=31$ .

Step 3. Substitute into $x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)$ : the bracket is $38-31=7$ , so $x^3+y^3+z^3-75 = 10\times7 = 70$ .

Step 4. Solve: $x^3+y^3+z^3 = 70+75 = 145$ . $x^3+y^3+z^3 = \mathbf{145}$ . Timing note: this chains two identities and should take three to four minutes. The first minute is spent recognising the three-variable identity needs $xy+yz+zx$ , which is missing; the rest is the two substitutions. Do not attempt to solve for $x, y, z$ individually — with only three symmetric equations, individual values are hard to pin down, but any symmetric combination like the sum of cubes follows directly from the identities.

Common traps

Dropping the middle term. Writing $(a+b)^2=a^2+b^2$ is the single most common error; the identity always has three terms when $a,b\ne0$ .
Sign slips. $(a-b)^2$ is not $a^2-b^2$ — that belongs to the different identity $(a+b)(a-b)$ , whose middle terms cancel rather than combine.
Guessing instead of checking both conditions. Splitting the middle term of $x^2+px+q$ needs both $a+b=p$ and $ab=q$ at once; matching only one gives wrong factors.
Forgetting a hidden common factor. $50p^2+60pq+18q^2$ will not match a perfect-square pattern until the factor $2$ is removed first.
Misapplying the difference-of-squares identity. $(2x+3)(2x-5)$ is not $(a+b)(a-b)$ , since the brackets do not share the same second term.
Cancelling terms instead of factors. In $\dfrac{x-4}{5(x+5)}$ , only a fully factored common bracket may cancel — never a lone term buried inside a sum.
Losing the sign pattern in cube identities. $(a-b)^3=a^3-3a^2b+3ab^2-b^3$ alternates in sign; copying $(a+b)^3$ 's all-positive pattern is a frequent slip.

Common trapThe most common mistake in this chapter is writing (a + b) squared as a squared + b squared, silently dropping the cross term 2ab

This single error costs more marks in Class 9 algebra than any other, and it resurfaces in disguise — students who correctly remember $(a+b)^2=a^2+2ab+b^2$ often still forget the analogous middle terms in $(a+b+c)^2$ or $(a+b)^3$ . Every one of these identities was proved using the distributive property, which guarantees cross terms between different pieces of a sum always appear — they are never optional and never vanish just because the expression looks tidy without them. Fast self-check: if squaring a two-term sum ever gives you back only two terms, stop and redo it. The same discipline catches the sign error in $(a-b)^2$ , which keeps all three terms but flips the middle sign, and never becomes $a^2-b^2$ .

You should be able to…

State the difference between an algebraic identity and an ordinary equation, with an example of each.
Prove $(a\pm b)^2=a^2\pm2ab+b^2$ both geometrically, by dissecting a square, and algebraically, using the distributive property.
Derive $(a+b+c)^2$ from the two-term square identity by treating two of the three terms as a single block.
Use the difference-of-squares identity, and its rearrangement $a^2=(a+b)(a-b)+b^2$ , to square numbers near a round base mentally.
Factorise a perfect-square quadratic by matching it to $a^2\pm2ab+b^2$ , removing a common factor first if needed.
Factorise a general quadratic $x^2+px+q$ by splitting the middle term, finding two numbers whose sum is $p$ and product is $q$ .
Visualise a factorisation as algebra tiles rearranged into a rectangle, and explain why only one split of the middle term tiles cleanly.
State and derive the cube identities $(a\pm b)^3$ , the sum and difference of cubes, and the three-variable identity for $x^3+y^3+z^3-3xyz$ .
Simplify a rational algebraic expression by factorising fully and cancelling only genuine common factors, never individual terms.
Recognise which identity's pattern an unfamiliar expression matches before expanding, factoring or simplifying it.

Ask the tutor: Want to work through why the difference-of-squares trick works for any two numbers near a common round base, or practice splitting the middle term on a few more quadratics until it feels automatic?

Why do we bother proving identities algebraically if the geometric picture with squares already looks convincing?

The geometric picture is a wonderful way to see why an identity like $(a+b)^2=a^2+2ab+b^2$ is true, but it quietly assumes $a$ and $b$ are positive lengths you can actually draw. Algebra makes no such assumption — the same proof, $(a+b)(a+b)=a(a+b)+b(a+b)=a^2+2ab+b^2$ , works exactly as well when $a$ and $b$ are negative, zero or fractions, none of which can be drawn as a side length. This chapter shows both routes for that reason: the picture builds intuition, while the algebra supplies the actual proof, valid without exception. When in doubt, trust the algebra, not the picture.

How do I know which identity to use when a problem does not tell me directly?

Look at the shape of what you are given, not the numbers in it. Two terms squared with a plus sign means $(a+b)^2$ ; a minus sign means $(a-b)^2$ ; three terms squared means the trinomial identity; two brackets sharing identical terms with opposite signs means the difference-of-squares pattern; and a quadratic matching no perfect-square pattern falls back to splitting the middle term, which always works for factorable quadratics with integer roots. With practice, matching the shape becomes automatic — until then, write the general pattern beside your expression and check term by term whether it fits.

Why can I cancel the (x minus 4) in a fraction like (x minus 4)(x minus 3) over 5(x minus 4)(x plus 5), but I cannot cancel an x on its own?

Cancelling in a fraction is really dividing both the numerator and the denominator by the same nonzero quantity, and that quantity must be a genuine common factor — something multiplied, not added. In $\dfrac{(x-4)(x-3)}{5(x-4)(x+5)}$ , the bracket $(x-4)$ multiplies everything else top and bottom, so dividing both by it, when nonzero, is valid and leaves $\dfrac{x-3}{5(x+5)}$ . But in $\dfrac{x+3}{x}$ , the $x$ in the denominator is not a factor of the whole numerator $x+3$ — it only touches one of its two terms — so nothing legitimate can cancel, and treating $\dfrac{x+3}{x}$ as $3$ is simply wrong.

Do I need to memorise all these identities separately, or is there a smaller set that generates the rest?

You genuinely only need to be fluent in a handful: the two square identities $(a\pm b)^2$ , the trinomial square $(a+b+c)^2$ , the difference of squares $a^2-b^2$ , the two cube identities $(a\pm b)^3$ , the sum and difference of cubes, and the three-variable cube identity. Everything else — squaring numbers mentally, factorising perfect squares, simplifying rational expressions — is an application of these, not a separate rule. If you forget one under exam pressure, rebuild it in under a minute using the distributive property, exactly as the two derivations above demonstrate — memorising the method is as valuable as memorising the formula.