Algebraic identities

Multiply out (2a+3b−c)2(2a + 3b - c)^2 term by term and you will write nine products, collect six of them into pairs, and probably drop a sign somewhere along the way. Or you can recognise the shape, reach for a memorised pattern, and write the answer in one line. That pattern is an algebraic identity — an equation that is true for every value of the variables, not just for some special ones, and that can be proven once by direct expansion and then reused forever. This is the closing topic of the CBSE (NCERT) Class 9 chapter Polynomials, sitting right after factorisation using the Factor Theorem. The syllabus asks for four standard identities: the square of a trinomial, the cube of a binomial sum or difference, the sum and difference of two cubes, and the three-variable identity for x3+y3+z3−3xyzx^3+y^3+z^3-3xyz. This article recalls each one, proves the key identity by expansion, and works through the exam traps that trip up nearly everyone the first time.

The core insight

An algebraic identity holds for all values of its variables and can always be verified by expanding the left-hand side with the distributive law, which is exactly why (x+y+z)2(x+y+z)^2, (x±y)3(x\pm y)^3, x3±y3x^3\pm y^3, and x3+y3+z3−3xyzx^3+y^3+z^3-3xyz can be trusted as instant expansion and factorisation shortcuts.

The core idea

An identity is different from an ordinary equation. An equation like x+3=7x + 3 = 7 is true only for one value of xx; an identity like (x+1)2=x2+2x+1(x+1)^2 = x^2+2x+1 is true for every value of xx you substitute. That is what makes identities safe to use as ready-made shortcuts instead of multiplying things out from scratch each time. The CBSE Class 9 syllabus fixes four identities to know cold. The first extends the familiar two-term square to three terms:

(x+y+z)2=x2+y2+z2+2xy+2yz+2zx(x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx

where: x,y,zx, y, z are any three terms (numbers, monomials, or expressions with a sign already attached). The next pair extends squaring to cubing, for a binomial sum and a binomial difference:

(x+y)3=x3+y3+3xy(x+y),(x−y)3=x3−y3−3xy(x−y)(x + y)^3 = x^3 + y^3 + 3xy(x + y), \qquad (x - y)^3 = x^3 - y^3 - 3xy(x - y)

where: the bracketed factor (x+y)(x+y) or (x−y)(x-y) on the right is the same binomial being cubed on the left — it is not re-expanded, just carried along. The third pair factorises a sum or a difference of two perfect cubes:

x3+y3=(x+y)(x2−xy+y2),x3−y3=(x−y)(x2+xy+y2)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 quadratic factor is not (x±y)2(x\pm y)^2 — the middle term's sign is deliberately flipped relative to the linear factor, which is exactly where most sign errors happen. The fourth is a genuinely three-variable identity, and it is the one this section's derivation proves in full below:

x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3 + y^3 + z^3 - 3xyz = (x + y + z)(x^2 + y^2 + z^2 - xy - yz - zx)

where: x,y,zx, y, z are any three terms, and the right-hand side splits neatly into a linear factor and a quadratic factor, mirroring the sum-of-cubes pattern but with an extra variable. A student who forgets a sign should not panic and guess — every one of these can be re-derived on the spot by multiplying the right-hand side back out with the distributive law, which is the whole point of calling them identities rather than facts to memorise blind. Each identity also runs in reverse: read left to right it is an expansion shortcut, and read right to left it is a factorisation shortcut, so the same four patterns cover both directions of the exam paper.

Figure
Four boxes showing the square of a trinomial the cube of a binomial the sum of cubes and the three variable identity as a quick visual reference card
The four Class 9 algebraic identities, at a glance (x + y + z)^2 = x^2 + y^2 + z^2 + 2xy + 2yz + 2zx square of a trinomial — six terms (x plus-or-minus y)^3 = x^3 plus-or-minus y^3 plus-or-minus 3xy(x plus-or-minus y) cube of a binomial — watch the middle sign x^3 plus-or-minus y^3 = (x plus-or-minus y)(x^2 minus-or-plus xy + y^2) sum or difference of cubes x^3+y^3+z^3-3xyz = (x+y+z)(x^2+y^2+z^2-xy-yz-zx) special case: if x+y+z=0, this equals 3xyz
Derivation AProving the three-cube identity by expanding the trinomial factor

The goal is to show that the right-hand side product (x+y+z)(x2+y2+z2−xy−yz−zx)(x+y+z)(x^2+y^2+z^2-xy-yz-zx) genuinely equals x3+y3+z3−3xyzx^3+y^3+z^3-3xyz when multiplied out term by term, so that the identity is not a fact to trust blindly but a result anyone can re-derive from the distributive law in under a minute during an exam.

Step 1. Distribute xx across the quadratic factor: x(x2+y2+z2−xy−yz−zx)=x3+xy2+xz2−x2y−xyz−x2zx(x^2+y^2+z^2-xy-yz-zx) = x^3 + xy^2 + xz^2 - x^2y - xyz - x^2z.

Step 2. Distribute yy across the same quadratic factor: y(x2+y2+z2−xy−yz−zx)=x2y+y3+yz2−xy2−y2z−xyzy(x^2+y^2+z^2-xy-yz-zx) = x^2y + y^3 + yz^2 - xy^2 - y^2z - xyz.

Step 3. Distribute zz across the same quadratic factor: z(x2+y2+z2−xy−yz−zx)=x2z+y2z+z3−xyz−yz2−xz2z(x^2+y^2+z^2-xy-yz-zx) = x^2z + y^2z + z^3 - xyz - yz^2 - xz^2.

Step 4. Add the three expansions from Steps 1 to 3 and cancel matching positive-negative pairs. The terms x2yx^2y and −x2y-x^2y cancel, xy2xy^2 and −xy2-xy^2 cancel, x2zx^2z and −x2z-x^2z cancel, xz2xz^2 and −xz2-xz^2 cancel, y2zy^2z and −y2z-y^2z cancel, and yz2yz^2 and −yz2-yz^2 cancel — six pairs of terms vanish completely.

Step 5. What survives is x3+y3+z3x^3 + y^3 + z^3 from the diagonal terms, plus three copies of −xyz-xyz (one from each step), giving x3+y3+z3−3xyzx^3+y^3+z^3-3xyz exactly, which matches the left-hand side and completes the proof.

Step 6. A direct consequence: if x+y+z=0x+y+z=0, the linear factor on the right is zero, so the whole right-hand side is zero, meaning x3+y3+z3−3xyz=0x^3+y^3+z^3-3xyz=0, i.e. x3+y3+z3=3xyzx^3+y^3+z^3=3xyz — a shortcut valid only under that one condition.

Worked example

Question. Use a suitable identity to expand (2a+3b−c)2(2a+3b-c)^2, and separately evaluate x3+y3+z3x^3+y^3+z^3 for x=4x=4, y=−3y=-3, z=−1z=-1 without cubing any number directly.

Step 1. For the first part, match (2a+3b−c)2(2a+3b-c)^2 to the pattern (x+y+z)2(x+y+z)^2 with x=2ax=2a, y=3by=3b, z=−cz=-c.

Step 2. Compute the three squared terms: x2=(2a)2=4a2x^2=(2a)^2=4a^2, y2=(3b)2=9b2y^2=(3b)^2=9b^2, z2=(−c)2=c2z^2=(-c)^2=c^2.

Step 3. Compute the three cross terms: 2xy=2(2a)(3b)=12ab2xy=2(2a)(3b)=12ab, 2yz=2(3b)(−c)=−6bc2yz=2(3b)(-c)=-6bc, 2zx=2(−c)(2a)=−4ac2zx=2(-c)(2a)=-4ac.

Step 4. Add all six terms to get (2a+3b−c)2=4a2+9b2+c2+12ab−6bc−4ac(2a+3b-c)^2 = 4a^2+9b^2+c^2+12ab-6bc-4ac, matching the Pass-1 result exactly.

Step 5. For the second part, check whether x+y+z=0x+y+z=0: here 4+(−3)+(−1)=04+(-3)+(-1)=0, so the special case applies.

Step 6. Since x+y+z=0x+y+z=0, use x3+y3+z3=3xyz=3(4)(−3)(−1)=3×12=∗∗36∗∗x^3+y^3+z^3=3xyz=3(4)(-3)(-1)=3\times12=**36**, obtained without cubing 44, −3-3, or −1-1 individually.

Competitive stretch

The same trick handles x3+y3+z3x^3+y^3+z^3 whenever the three numbers happen to sum to zero, which is a common disguise in speed-math and olympiad-style problems — always test x+y+z=0x+y+z=0 before reaching for a calculator to cube anything.

Common traps

  • Dropping the extra minus sign in (x−y)3=x3−y3−3xy(x−y)(x-y)^3 = x^3-y^3-3xy(x-y) — the middle term keeps a negative sign that is easy to lose.
  • Writing the quadratic factor of x3+y3x^3+y^3 as x2+xy+y2x^2+xy+y^2 instead of x2−xy+y2x^2-xy+y^2 (and making the mirror-image mistake for x3−y3x^3-y^3).
  • Applying the shortcut x3+y3+z3=3xyzx^3+y^3+z^3=3xyz without first checking x+y+z=0x+y+z=0 — the shortcut is only valid under that condition, never in general.
  • Treating identities as facts to memorise rather than results that can be re-derived by expanding the right-hand side, which is the safest way to recover a forgotten sign under exam pressure.

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Why does the sign flip between (x+y)^3 and (x-y)^3 but not in the same way as (x-y)^2?

Squaring always produces a positive middle term because a negative times a negative is positive, so (x−y)2=x2−2xy+y2(x-y)^2=x^2-2xy+y^2 has just one sign change from (x+y)2(x+y)^2. Cubing keeps odd powers of the negative sign, so (x−y)3(x-y)^3 picks up minus signs on both the y3y^3 term and the 3xy(x−y)3xy(x-y) term. The safest fix is to re-derive it: substitute −y-y for yy in (x+y)3(x+y)^3 and simplify carefully rather than guessing which signs flip.

Do these identities work if x, y, or z are themselves expressions, not just single letters?

Yes — an identity holds for any values substituted in, including whole expressions. For example, in (2a+3b−c)2(2a+3b-c)^2, the roles of xx, yy, zz are played by 2a2a, 3b3b, and −c-c respectively, each of which is itself a term with a coefficient. Just be careful to square and multiply the entire substituted term, including its sign and coefficient, not just the letter part.

Is there a way to remember when x^3+y^3+z^3=3xyz applies without memorising it as a separate rule?

Yes — treat it as a consequence, not a separate identity. Start from x3+y3+z3−3xyz=(x+y+z)(x2+y2+z2−xy−yz−zx)x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx), and notice that if the first bracket (x+y+z)(x+y+z) is zero, the entire right-hand side becomes zero, which forces x3+y3+z3−3xyz=0x^3+y^3+z^3-3xyz=0. Rearranging gives x3+y3+z3=3xyzx^3+y^3+z^3=3xyz. Deriving it this way means you can never apply it under the wrong condition, because you always check x+y+z=0x+y+z=0 first.

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