Algebraic identities

Multiplying $103 \times 103$ the long way takes a page of working — unless you notice that $103 = 100+3$ and reach for a shortcut you already half-know: $(a+b)^2 = a^2+2ab+b^2$ . That single line turns a messy multiplication into three easy terms added up in your head. An algebraic identity is an equation that is true for every value of the letters in it, not just one special value, and the ICSE Class 8 chapter gives you a small toolkit of these — four square and product identities, plus the cube identities that follow the same logic with one more factor. Learn to recognise which identity a given expression matches, and expansion, mental squaring, and evaluating expressions from $a+b$ and $ab$ all become one-line substitutions instead of slow multiplication.

The core insight

A handful of standard identities — proved once by expanding brackets with the distributive law — let you expand, square, cube, or evaluate algebraic expressions instantly by substitution instead of multiplying term by term every time.

The core idea

An identity differs from an ordinary equation. The equation $2x+3=7$ is true only when $x=2$ ; the identity $(a+b)^2=a^2+2ab+b^2$ is true for every value you put in for $a$ and $b$ . That is what makes identities reusable tools rather than one-off facts to solve.

The four standard square/product identities. These are the core of the ICSE Class 8 chapter, and every one of them follows from multiplying out brackets using the distributive law you already know:

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

(a+b)(a-b) = a^2-b^2 \qquad (x+a)(x+b) = x^2+(a+b)x+ab

where: $a$ , $b$ , $x$ stand for any numbers, monomials, or algebraic terms; $2ab$ is the middle "cross term" that appears when you multiply two binomials; $a^2-b^2$ is called the difference of two squares.

The cube identities. Following the standard Class 8 syllabus, the same idea extends one factor further, to $(a+b)^3$ and the sum/difference of cubes:

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

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

where: the coefficients $1,3,3,1$ in the cube expansions are the same pattern you would get from repeated multiplication; the powers of $a$ fall from 3 to 0 while the powers of $b$ rise from 0 to 3, which is a memory aid, not a coincidence.

How the identities connect. $(a+b)^2$ and $(a-b)^2$ differ only in the sign of the middle term, because swapping $b$ for $-b$ throughout the first identity produces the second. And $(a+b)(a-b)$ is really the $(x+p)(x+q)$ pattern with $p=b$ , $q=-b$ : the middle coefficient becomes $p+q=b+(-b)=0$ , which is exactly why the middle term vanishes and only $a^2-b^2$ survives. Four separate-looking facts collapse into one family once you see this.

Why bother memorising these at all? Every one of the eight identities above can be reconstructed from scratch just by multiplying out the brackets with the distributive law — nothing here is a magic rule handed down without reason. The point of memorising them is speed: in an exam, expanding $(3x-5y)^2$ term by term from first principles takes far longer than recognising the pattern $a=3x$ , $b=5y$ and writing down $9x^2-30xy+25y^2$ in one line. Treat the identities the way you treat multiplication tables — facts you could always re-derive, but that you keep ready-to-hand because re-deriving them every single time wastes valuable exam minutes. The same logic applies in the Factorisation chapter that follows this one, where you will run every one of these identities in reverse: given an expanded polynomial, you spot which identity produced it and factorise instantly instead of guessing.

Figure

A square of side a plus b split into four regions a squared, a times b, a times b, and b squared showing the geometric meaning of the identity a plus b whole squared

Derivation of the (a+b)^2 identity

Derivation AProving (a+b)^2 = a^2 + 2ab + b^2 from the distributive law

The goal is to show that the identity $(a+b)^2 = a^2+2ab+b^2$ is not a rule to memorise blindly but a direct, guaranteed consequence of the distributive law of multiplication over addition applied twice, which means it must hold true for every possible value of $a$ and $b$ without exception, including negative numbers, fractions, and algebraic terms.

Step 1. Rewrite the square as a product. By definition, squaring means multiplying a quantity by itself, so $(a+b)^2$ means $(a+b)\times(a+b)$ . Write it as a product of two brackets: $(a+b)(a+b)$ .

Step 2. Apply the distributive law to the first bracket. Treat the first $(a+b)$ as a single multiplier and distribute it over the two terms of the second bracket: $(a+b)(a+b) = a(a+b) + b(a+b)$ .

Step 3. Distribute again inside each piece. $a(a+b) = a\cdot a + a\cdot b = a^2+ab$ , and $b(a+b) = b\cdot a + b\cdot b = ab+b^2$ . Putting the two pieces together gives $(a+b)^2 = a^2+ab+ab+b^2$ .

Step 4. Combine the like terms. The two middle terms $ab$ and $ab$ are identical, so they add to $2ab$ , leaving $(a+b)^2 = a^2+2ab+b^2$ exactly as claimed. Because every step used only the distributive law, which holds for all real numbers, the result holds for every choice of $a$ and $b$ — that is precisely what makes it an identity rather than an equation to be solved.

Worked example

Find $103^2$ using an identity, then use the same identity to evaluate $a^2+b^2$ when $a+b=9$ and $ab=14$ .

Step 1. Write $103$ as $(100+3)$ , matching the pattern $(a+b)^2$ with $a=100$ and $b=3$ .

Step 2. Apply $(a+b)^2=a^2+2ab+b^2$ : $103^2 = 100^2+2(100)(3)+3^2 = 10000+600+9$ .

Step 3. Add the terms: $103^2 = 10609$ .

Step 4. For the second part, rearrange the same identity: $a^2+b^2=(a+b)^2-2ab$ . Substituting $a+b=9$ and $ab=14$ gives $a^2+b^2 = 9^2-2(14) = 81-28 = \mathbf{53}$ .

Competitive stretch

The same rearrangement finds $a-b$ too, using $(a-b)^2=(a^2+b^2)-2ab = 53-28=25$ , so $a-b=\sqrt{25}=5$ (taking $a>b$ ). This matches $a=7,b=2$ since $7+2=9$ and $7\times2=14$ . The same identity, read as $(a+b)(a-b)=a^2-b^2$ , also speeds up products like $52\times48=(50+2)(50-2)=50^2-2^2=2500-4=2496$ whenever two numbers sit equally spaced around a round number. The cube identities extend this same mental-maths trick one step further. To cube $99$ , write $99=(100-1)$ , matching $(a-b)^3$ with $a=100$ , $b=1$ : $(100-1)^3 = 100^3-3(100)^2(1)+3(100)(1)^2-1^3 = 1000000-30000+300-1 = 970299$ . Notice how each term is computed independently before the final addition, exactly as in the square-identity example above — the pattern of substituting into a memorised formula, rather than multiplying $99\times99\times99$ directly, is what makes both tricks fast under exam conditions.

Common traps

Writing $(a+b)^2=a^2+b^2$ , dropping the middle term $2ab$ entirely — the single most common error in this chapter.
Mixing up the sign in $(a-b)^2=a^2-2ab+b^2$ , writing $-2ab$ as $+2ab$ or vice-versa for $(a+b)^2$ .
Confusing $(a+b)(a-b)=a^2-b^2$ with $(a-b)^2$ : these expand to very different results even though the brackets look similar.
Applying a square identity where a cube is needed — $(a+b)^3$ has four terms, not the three terms of $(a+b)^2$ , so mixing the two formulas gives a wrong answer.

Ask the tutor: Not sure which identity fits a particular expression, or want more practice evaluating expressions from $a+b$ and $ab$ ? Ask the AI tutor to work through more examples with you.

Why is (a+b)^2 not just a^2 + b^2? It feels like squaring should just square each term.

Squaring a sum is not the same as squaring each part separately, because $(a+b)^2$ means $(a+b)$ multiplied by itself, and multiplying two binomials always produces a middle cross term. Try it with numbers: $(2+3)^2=25$ , but $2^2+3^2=13$ — clearly different. The missing piece is $2ab$ , which comes from the two cross-multiplications $a\times b$ and $b\times a$ in the distributive expansion. Always include it.

Do I really need to memorise the cube identities, or can I just multiply it out each time?

You can always multiply $(a+b)^3$ as $(a+b)^2\times(a+b)$ if you forget the formula, and doing this once is a good way to convince yourself the identity is correct. But under exam time pressure, multiplying out four terms takes far longer than substituting into a memorised identity. Notice the coefficient pattern $1,3,3,1$ and the powers of $a$ falling while powers of $b$ rise — that pattern makes the formula much easier to recall than treating it as four unrelated terms.

How do I decide which identity to use when I see a new expression?

Look at the structure before you touch the algebra. Two identical brackets being squared, like $(2x+3y)^2$ , is the $(a+b)^2$ or $(a-b)^2$ pattern depending on the sign. Two brackets with the same terms but opposite signs, like $(2x+3y)(2x-3y)$ , is the difference-of-squares pattern. Two brackets sharing the same first letter but different constants, like $(x+5)(x+3)$ , is the $(x+a)(x+b)$ pattern where the middle coefficient is the sum $a+b$ and the constant is the product $ab$ . Matching the shape first saves you from expanding blindly.