Algebra

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

⏱ 3 min read🧩 5 question types🎯 13 practice Q

The idea in one minute

The three-variable cube identity $a^3+b^3+c^3-3abc = (a+b+c)(\dots)$ has two money forms: a zero sum collapses the cubes to $3abc$ , and the equal-squares condition forces $a=b=c$ . Everything else routes through the power-sum expansion.

Overview

The master identity:

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

Rewrite the bracket with the sum alone, using $a^2+b^2+c^2 = s^2 - 2P$ :

a^3+b^3+c^3-3abc = s\left(s^2 - 3P\right) \quad \text{where } s = a+b+c,\ P = ab+bc+ca

With $s = 6$ and $P = 11$ : the value is $6 \times (36 - 33) = 18$ . One substitution, no cubing.

The zero-sum shortcut

If $s = 0$ , the whole right side vanishes except one reading: $a^3+b^3+c^3 = 3abc$ . The exam hides the zero sum inside brackets: $(x-y) + (y-z) + (z-x) = 0$ always. So $(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$ , instantly. Check the shape before expanding anything: with $x-y = 2$ and $y-z = 3$ , the third bracket is $-5$ , and $3 \times 2 \times 3 \times (-5) = -90$ .

Rule: Add the three terms first. If they sum to zero, the cube-sum is three times the product, signs included.

Zero sum, squares and fourth powers

When $s = 0$ , squaring the sum gives $a^2+b^2+c^2 = -2P$ . Square once more for fourth powers: $a^4+b^4+c^4 = 2P^2$ . Check with $1, 1, -2$ : the squares give $6 = -2 \times (-3)$ , and the fourth powers give $18 = 2 \times 9$ . These two lines answer every 'if $a+b+c=0$ ' stem that asks about squares or fourth powers, no cubes needed.

The equal case

The bracket also equals $\dfrac{1}{2}\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$ , which is never negative and is zero exactly when $a = b = c$ . So the condition $a^2+b^2+c^2 = ab+bc+ca$ forces all three equal. With the extra fact $x+y+z = 12$ , each variable is $4$ , and quantities like $xyz = 64$ follow at once.

Power sums

The cube of the sum expands as

a^3+b^3+c^3 = s^3 - 3sP + 3R \quad (R = abc)

Given any three of $\{s, P, R, \text{cube-sum}\}$ , the fourth is one linear step. Example: $s = 4$ , $R = 2$ , cube-sum $= 10$ (the numbers are $1, 1, 2$ ). Then $64 = 10 + 12P - 6$ , so $P = 5$ .

Tip: A useful companion: $(a+b)(b+c)(c+a) = sP - R$ . Check with $1, 2, 3$ : $3 \times 5 \times 4 = 60$ and $6 \times 11 - 6 = 60$ .

Close numbers: the half form

When the three numbers sit close together, $\dfrac{1}{2}(a+b+c)\left[(a-b)^2+(b-c)^2+(c-a)^2\right]$ is tiny arithmetic. For $25^3+24^3+23^3 - 3 \times 25 \times 24 \times 23$ : half of $72$ times $(1 + 1 + 4)$ gives $36 \times 6 = 216$ .

An order that works

Do the three brackets or variables sum to zero? Use $3 \times$ product.
Are $s$ and $P$ given? Use $s(s^2 - 3P)$ .
Is $abc$ in the question? Use the power-sum expansion.
Does $a^2+b^2+c^2 = ab+bc+ca$ hold? All variables are equal.

Question types you will see

Each type: how to recognise it, the method step by step, and one question to try.

Type 1very common3 practice Q

Zero-sum cubes

How to spot it:

Three cubes whose bases or brackets add to zero, or a stated condition $a+b+c=0$ with $abc$ known.

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

Method

Add the three bases or brackets; confirm zero.
Multiply the three together.
Triple the product, watching signs.

Why it works:

The master identity carries the factor $(a+b+c)$ , and zero kills the rest.

Try this

If $a + b + c = 0$ and $abc = -20$ , find $a^3 + b^3 + c^3$ .

Show solution

The sum is zero, so the rule applies.
$3 \times abc = 3 \times (-20)$ .
$= -60$ .

Answer

-60

Type 2very common3 practice Q

The sum form s times s squared minus 3P

How to spot it:

The sum $a+b+c$ and the pairwise sum $ab+bc+ca$ are given; $a^3+b^3+c^3-3abc$ is asked.

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

Method

Name $s = a+b+c$ and $P = ab+bc+ca$ .
Compute $s^2 - 3P$ .
Multiply by $s$ .

Why it works:

Substituting $a^2+b^2+c^2 = s^2 - 2P$ leaves the bracket as $s^2 - 3P$ .

Try this

If $a + b + c = 9$ and $ab + bc + ca = 26$ , find $a^3+b^3+c^3-3abc$ .

Show solution

$s^2 = 81$ .
$3P = 78$ , so $s^2 - 3P = 3$ .
$9 \times 3 = 27$ .

Answer

Type 3common2 practice Q

The equal variables case

How to spot it:

The condition $x^2+y^2+z^2 = xy+yz+zx$ appears, and a symmetric quantity is asked.

x^2+y^2+z^2 = xy+yz+zx \Rightarrow x = y = z

Method

Recognise the condition; it forces all three equal.
Find the common value from the given sum: divide by 3.
Evaluate the asked quantity with the common value.

Why it works:

The difference of the two sides is half a sum of squares, zero only when all equal.

Try this

If $x^2+y^2+z^2 = xy+yz+zx$ and $x+y+z = 12$ , find $xyz$ .

Show solution

The condition forces $x = y = z$ .
$x + y + z = 12$ gives each variable $4$ .
$xyz = 4 \times 4 \times 4 = 64$ .

Answer

Type 4common2 practice Q

Power sums with the product

How to spot it:

Three of sum, pairwise sum, product and cube-sum are given; the fourth is asked.

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

Method

Write the expansion with the known letters in place.
Substitute the three known quantities.
Solve the resulting linear equation for the fourth.

Why it works:

The cube of the sum expands into exactly these four symmetric pieces.

Try this

If $a + b + c = 4$ , $abc = 2$ and $a^3+b^3+c^3 = 10$ , find $ab+bc+ca$ .

Show solution

$s^3 = 64$ .
$64 = 10 + 3 \times 4 \times P - 3 \times 2$ .
$12P = 60$ , so $P = 5$ (the numbers are $1, 1, 2$ ).

Answer

Type 5common

Pair products from s, P and R

How to spot it:

The question asks for $(a+b)(b+c)(c+a)$ while $a+b+c$ , $ab+bc+ca$ and $abc$ are given or findable.

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

Method

Collect $s = a+b+c$ , $P = ab+bc+ca$ and $R = abc$ .
Substitute into $sP - R$ .
Keep the product in one line; expand nothing.

Why it works:

Each pair sum is the total minus the missing variable, and the three multiply to exactly $sP - R$ .

Try this

If $a + b + c = 10$ , $ab + bc + ca = 27$ and $abc = 18$ , find $(a+b)(b+c)(c+a)$ .

Show solution

$sP = 10 \times 27 = 270$ .
$sP - R = 270 - 18 = 252$ .
Check with $1, 3, 6$ : $4 \times 9 \times 7 = 252$ .

Answer

252

Formula sheet

Master identity

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

Sum form

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

Half form

a^3+b^3+c^3-3abc = \tfrac12\,s\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 case

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

Power-sum expansion

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

Pair products

(a+b)(b+c)(c+a) = sP - R

Pairwise from squares

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

Shortcuts that save time

⚡ Hunt for a hidden zero sum

Brackets like x minus y, y minus z, z minus x always sum to zero. The cube-sum is three times the product.

Example

If x - y = 2 and y - z = 3, find (x-y)^3 + (y-z)^3 + (z-x)^3.

Show solution

$z - x = -(2 + 3) = -5$ .
The three brackets sum to zero, so the cube-sum is $3$ times the product.
$3 \times 2 \times 3 \times (-5) = -90$ .

Answer

-90

⚡ Close numbers: use the half form

When the three numbers differ by little, the squared differences are tiny, and half of s times their sum is quick arithmetic.

Example

Find 25^3 + 24^3 + 23^3 - 3 x 25 x 24 x 23.

Show solution

$s = 72$ ; differences: $(a-b)^2 = 1$ , $(b-c)^2 = 1$ , $(c-a)^2 = 4$ .
$\dfrac{1}{2} \times 72 \times 6$ .
$= 216$ .

Answer

216

⚡ Value-putting under a condition

If a condition like a+b+c=0 is given, pick simple numbers that satisfy it and evaluate the expression.

Example

If a + b + c = 0, find a^2/bc + b^2/ca + c^2/ab.

Show solution

Put $a = 1$ , $b = 1$ , $c = -2$ .
$\dfrac{1}{-2} + \dfrac{1}{-2} + \dfrac{4}{1}$ .
$= -\dfrac{1}{2} - \dfrac{1}{2} + 4 = 3$ .

Answer

Mistakes to avoid

Where most students lose marks on this subtopic.

Mistake 01

Using $a^3+b^3+c^3 = 3abc$ without checking the sum is zero.

The shortcut is valid only when $a+b+c = 0$ .

Mistake 02

Dropping the half in the half form.

The factor is one half of $s$ times the squared differences; losing it doubles the answer.

Mistake 03

Forgetting the $+3abc$ term in the power-sum expansion.

$s^3$ expands to cube-sum plus $3sP$ minus $3R$ ; keep every sign.

Mistake 04

Losing signs with negative variables.

Two negatives multiply to a positive; recount minus signs before the final multiply.

Quick revision

Read this the night before the exam.

$a^3+b^3+c^3-3abc = s(s^2 - 3P)$ .
$a+b+c = 0$ gives $a^3+b^3+c^3 = 3abc$ .
$(x-y)^3+(y-z)^3+(z-x)^3 = 3(x-y)(y-z)(z-x)$ .
$a^2+b^2+c^2 = ab+bc+ca$ forces $a = b = c$ .
$a^3+b^3+c^3 = s^3 - 3sP + 3R$ .
$(a+b)(b+c)(c+a) = sP - R$ .

Practice: 13 questions

Sets of 10, mixed across the question types above. Every answer has a step-by-step explanation.

Topic test · 13 questions

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

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