Algebra

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high importance~3 Q in Tier 143 formulas⚡ 21 shortcuts6 subtopics

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Subtopic 3 of 6·← $x+\frac{1}{x}$ type expressions·Surds: rationalisation and square roots of surds →

$a^3+b^3+c^3-3abc$ and conditional identities

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The three-variable cube identity is a CGL favourite because it has two useful forms and a powerful special case.

If $a+b+c=0$ , then $a^3+b^3+c^3=3abc$ . Questions hide the zero sum, e.g. $(x-y)+(y-z)+(z-x)=0$ .
If $a^2+b^2+c^2=ab+bc+ca$ , then $(a-b)^2+(b-c)^2+(c-a)^2=0$ , so $a=b=c$ .
For numbers that are close together (like 25, 24, 23), use the half-sum-of-squared-differences form.

📖 Notes 🧩 4 question types ⚡ Formulas 🎯 13 practice Q

Detailed notes

The identity and its two forms

$a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ The trinomial in brackets rewrites using the sum alone: $a^3+b^3+c^3-3abc = (a+b+c)\left[(a+b+c)^2 - 3(ab+bc+ca)\right]$ Everything the exam asks routes through these two forms. With $a+b+c = 6$ and $ab+bc+ca = 11$ : the value is $6\times(36-33) = 18$ .

The zero-sum shortcut

If $a+b+c = 0$ , the left factor vanishes and the identity collapses to $a^3+b^3+c^3 = 3abc$ This one shortcut answers a whole family: "if $a+b+c=0$ , $abc=8$ , find $a^3+b^3+c^3$ " → 24. It also explains why $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$ — the three terms sum to zero, so the cube-sum is 3×(product). Terms like $(b+c-3a)$ are engineered the same way: check whether the three brackets add to zero FIRST, before any expansion.

The equal case

$a^2+b^2+c^2-ab-bc-ca = \tfrac12\left[(a-b)^2+(b-c)^2+(c-a)^2\right] \ge 0$ So the bracket above is zero exactly when $a = b = c$ . Hence $a^2+b^2+c^2 = ab+bc+ca \iff a=b=c$ . If additionally $a+b+c$ is known, every symmetric expression follows: with sum 9 and that condition, $a=b=c=3$ and $a^2+b^2+c^2 = 27$ .

Power sums

The full expansion of the cube of the sum: $a^3+b^3+c^3 = (a+b+c)^3 - 3(a+b+c)(ab+bc+ca) + 3abc$ Given any three of the four quantities ( $a+b+c$ , $ab+bc+ca$ , $abc$ , $a^3+b^3+c^3$ ), the fourth drops out in one line. E.g. sum 9, pairwise 11, abc 6 → $729 - 297 + 18 = 450$ . A useful companion: $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca) - abc$ , and $(a+b+c)^3 = a^3+b^3+c^3 + 3(a+b)(b+c)(c+a)$ . The pair form is also how $(x+y+z)^3 - x^3-y^3-z^3 = 3(x+y)(y+z)(z+x)$ is proved — subtract the individual cubes from the expansion.

How to attack any question

Is a bracket-sum zero (like $x-y$ , $y-z$ , $z-x$ )? → cubes sum to 3×product.
Is $a+b+c$ given with $ab+bc+ca$ ? → factor/half form for $a^3+b^3+c^3-3abc$ .
Is $abc$ involved? → power-sum expansion.
Does $a^2+b^2+c^2$ equal $ab+bc+ca$ ? → all variables equal.

Common traps

Forgetting the $+3abc$ term in the power-sum expansion (the commonest algebra slip in CGL).
Expanding $(x-y)^3+(y-z)^3+(z-x)^3$ brute-force — it is 3×product, instantly.
Missing the half in $\frac12[(a-b)^2+\dots]$ when proving the equal case.
Sign errors when a variable is negative (e.g. $c=-6$ with $a=11$ , $b=-5$ ).

Quick revision

$a^3+b^3+c^3-3abc = (a+b+c)\left[(a+b+c)^2-3(ab+bc+ca)\right]$ .
$a+b+c=0 \Rightarrow a^3+b^3+c^3 = 3abc$ .
$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$ .
$a^2+b^2+c^2 = ab+bc+ca \iff a=b=c$ .
$a^3+b^3+c^3 = s^3 - 3sP + 3R$ where $s$ = sum, $P$ = pairwise, $R$ = product.

Types of questions asked

Every way this subtopic shows up in exams — how to recognise it, the formula or logic to use, and a solved example.

Type 1: Zero-sum cubesvery common3 practice Q

How to spot it:

A bracket sum of zero is hidden in the stem: $a+b+c=0$ , or terms like $(x-y)^3+(y-z)^3+(z-x)^3$ .

a+b+c=0 \Rightarrow a^3+b^3+c^3=3abc

Check whether the three terms (or the three variables) add to zero.
If yes, replace the cube-sum by 3×(product) — done in one line.
For $(x-y)^3+(y-z)^3+(z-x)^3$ the product is $(x-y)(y-z)(z-x)$ .

Why: with zero sum the big identity reduces to exactly this, for any three terms adding to zero.

Example: If $(x-y)=2$ and $(y-z)=3$ , then $(x-y)^3+(y-z)^3+(z-x)^3$ equals:

Brackets sum to 0 → cube-sum $= 3(x-y)(y-z)(z-x)$ . $z-x = -(2+3) = -5$ → value $= 3\times2\times3\times(-5) = -90$ .

Type 2: Factor / half form of the cube identityvery common3 practice Q

How to spot it:

$a+b+c$ and $ab+bc+ca$ (or $a^2+b^2+c^2$ ) given; $a^3+b^3+c^3-3abc$ asked.

a^3+b^3+c^3-3abc = s(s^2-3P)

Write $s = a+b+c$ , $P = ab+bc+ca$ .
If $a^2+b^2+c^2$ is given instead: $P = \frac{s^2-(a^2+b^2+c^2)}{2}$ .
Answer $= s(s^2-3P)$ .

Why: the bracket is exactly $(a+b+c)^2 - 3P$ after substituting $a^2+b^2+c^2 = s^2-2P$ .

Example: If $a+b+c=6$ and $ab+bc+ca=11$ , then $a^3+b^3+c^3-3abc$ equals:

$6\times(36 - 33) = 18$ .

Type 3: Equal variables casecommon2 practice Q

How to spot it:

The condition $a^2+b^2+c^2 = ab+bc+ca$ (or a rearranged version) appears; conclude all variables are equal.

a^2+b^2+c^2-ab-bc-ca = \tfrac12[(a-b)^2+(b-c)^2+(c-a)^2] = 0 \Rightarrow a=b=c

Recognise the half-sum-of-squares form; the condition forces $a=b=c$ .
Substitute the common value $= \frac{a+b+c}{3}$ .
Evaluate any symmetric expression from there.

Why: squares are non-negative, so their sum vanishes only pairwise.

Example: If $x^2+y^2+z^2 = xy+yz+zx$ and $x+y+z=9$ , then $x^2+y^2+z^2$ equals:

$x=y=z=3$ → $x^2+y^2+z^2 = 27$ .

Type 4: Power sums with abccommon2 practice Q

How to spot it:

Three of $\{a+b+c,\ ab+bc+ca,\ abc,\ a^3+b^3+c^3\}$ are given; the fourth is asked.

s^3 = a^3+b^3+c^3 + 3sP - 3R

Write the expansion $a^3+b^3+c^3 = s^3 - 3sP + 3R$ .
Substitute the three known quantities.
Solve for the fourth — one linear step, no factoring of cubics.

Why: the cube of the sum expands into exactly these symmetric pieces.

Example: If $a+b+c=6$ , $abc=6$ and $a^3+b^3+c^3=90$ , then $ab+bc+ca$ equals:

$216 = 90 + 18P - 18$ → $18P = 144$ → $P = 8$ .

Formulas

Main identity

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)

Half form

a^3+b^3+c^3-3abc=\tfrac12(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]

Zero-sum case

a+b+c=0\ \Rightarrow\ a^3+b^3+c^3=3abc

Equal-variables case

a^2+b^2+c^2=ab+bc+ca\ \Rightarrow\ a=b=c

Pairwise products

ab+bc+ca=\frac{(a+b+c)^2-(a^2+b^2+c^2)}{2}

Cyclic differences

(a-b)^3+(b-c)^3+(c-a)^3=3(a-b)(b-c)(c-a)

Shortcut tricks

⚡ Hunt for a hidden zero sum

$(x-y)+(y-z)+(z-x)=0$ always, so the sum of their cubes is $3(x-y)(y-z)(z-x)$ . Similarly $(a^2-b^2)+(b^2-c^2)+(c^2-a^2)=0$ .

Example: Find $\dfrac{(x-y)^3+(y-z)^3+(z-x)^3}{9(x-y)(y-z)(z-x)}$ .

Numerator $=3(x-y)(y-z)(z-x)$ , so the value is $\frac{3}{9}=\frac13$ .

⚡ Close numbers: use the half form

When $a,b,c$ differ by small amounts, $(a-b)^2+(b-c)^2+(c-a)^2$ is tiny and easy to compute.

Example: Find $25^3+24^3+23^3-3\times25\times24\times23$ .

$\tfrac12(72)\left[1^2+1^2+2^2\right]=36\times6=216$ .

⚡ Value-putting under a condition

If a condition like $a+b+c=0$ is given, pick numbers that satisfy it (e.g. $1,1,-2$ ) and evaluate.

Example: If $a+b+c=0$ , find $\dfrac{a^2}{bc}+\dfrac{b^2}{ca}+\dfrac{c^2}{ab}$ .

Put $a=1,b=1,c=-2$ : $\frac{1}{-2}+\frac{1}{-2}+\frac{4}{1}=3$ . (Algebraically it is $\frac{a^3+b^3+c^3}{abc}=\frac{3abc}{abc}=3$ .)

Where students lose marks

Using $a^3+b^3+c^3=3abc$ without first checking that $a+b+c=0$ .
Dropping the $\tfrac12$ in the half form, which doubles the answer.
Sign slips with negatives: for $a=11,b=-5,c=-6$ , $3abc=3(11)(-5)(-6)=+990$ .

Practice sets — 13 questions

Sets of 10, mixed across the question types above. Each answer comes with a step-by-step explanation.

Topic test · 13 questions

Suggested time 9 min · wrong answers go to your mistake notebook automatically.