MathsCBSE Class 9 · Exploring Algebraic Identities

Cubic Identities and the Sum of Three Cubes: Class 9

Q: Does $a^2 - ab + b^2$ factorise further?

Not over the real numbers. Its discriminant is negative, so it has no real linear factors.

Q: How do I remember the signs in the sum and difference of cubes?

Use SOAP. For a^3 + b^3 the linear factor is a + b (same), the middle term of the quadratic is -ab (opposite), and the last term +b^2 is always positive.

Q: Why does $a + b + c = 0$ give $a^3+b^3+c^3 = 3abc$?

Because a^3+b^3+c^3-3abc = (a+b+c)(\ldots), and the first factor is zero.

Updated 17 October 2026 · Fact-checked

The cube identities are $(a\pm b)^3 = a^3 \pm 3a^2b + 3ab^2 \pm b^3$, $a^3 + b^3 = (a+b)(a^2 - ab + b^2)$, $a^3 - b^3 = (a-b)(a^2 + ab + b^2)$ and $a^3 + b^3 + c^3 - 3abc = (a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. When $a + b + c = 0$ this gives $a^3 + b^3 + c^3 = 3abc$. Use them to expand, factorise and evaluate big cubes.

Understand Cubic Identities and the Sum of Three Cubes

Cube of a sum and difference. Multiplying $(a + b)^{2}$ by $(a + b)$ gives $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ . A geometric picture helps: a cube of edge $a + b$ splits into a cube of edge $a$ , a cube of edge $b$ , three slabs $a \times a \times b$ and three slabs $a \times b \times b$ . Replacing $b$ with $- b$ gives $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$ ; the signs alternate. A useful rearranged form is $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ and $a^{3} - b^{3} = (a - b)^{3} + 3 ab (a - b)$ , which turns "sum and product" data directly into cube values.

Sum and difference of cubes. $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ and $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$ . The sign in the bracket is the opposite of the sign outside for the middle term, and the quadratic factor does not factorise further over the real numbers. For $27 a^{3} - 8 b^{3}$ the cube roots are $3 a$ and $2 b$ , so the answer is $(3 a - 2 b) (9 a^{2} + 6 ab + 4 b^{2})$ .

The identity with three cubes. $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ . The second bracket equals $\frac{1}{2} [(a - b)^{2} + (b - c)^{2} + (c - a)^{2}]$ , which is zero only when $a = b = c$ . So the left side is $0$ exactly when $a + b + c = 0$ or $a = b = c$ . In particular, if $a + b + c = 0$ then $a^{3} + b^{3} + c^{3} = 3 ab c$ . This is how you evaluate $(- 12)^{3} + 7^{3} + 5^{3}$ without computing any cube: the three numbers add to zero, so the result is $3 \times (- 12) \times 7 \times 5 = - 1260$ .

Reciprocal pairs. From $x + \frac{1}{x} = m$ you get $x^{3} + \frac{1}{x ^{3}} = m^{3} - 3 m$ (because $x \cdot \frac{1}{x} = 1$ ). Similarly $x - \frac{1}{x} = n$ gives $x^{3} - \frac{1}{x ^{3}} = n^{3} + 3 n$ . These appear as 3-mark questions, often after finding $x^{2} + \frac{1}{x ^{2}}$ .

Numbers near a round number. $10 3^{3} = (100 + 3)^{3} = 1092727$ and $9 9^{3} = (100 - 1)^{3} = 970299$ . Write the three-term expansion with its binomial coefficients $1, 3, 3, 1$ and keep the powers organised in columns to avoid slips.

Exam habits. Always state which identity you use. When a number is a cube, say so (for example, $64 x^{3} = (4 x)^{3}$ ). For six-degree expressions such as $x^{6} - y^{6}$ , factor as a difference of squares first, then use both cube identities on $x^{3} - y^{3}$ and $x^{3} + y^{3}$ .

Cubic Identities and the Sum of Three Cubes: formulas to remember

Cube of a sum: $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$; Also $a^{3} + b^{3} + 3 ab (a + b)$ .
Cube of a difference: $(a - b)^{3} = a^{3} - 3 a^{2} b + 3 a b^{2} - b^{3}$; Signs alternate.
Sum of cubes: $a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$; Also $(a + b)^{3} - 3 ab (a + b)$ .
Difference of cubes: $a^{3} - b^{3} = (a - b) (a^{2} + ab + b^{2})$; Also $(a - b)^{3} + 3 ab (a - b)$ .
Three cubes: $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$; Zero when $a + b + c = 0$ or $a = b = c$ .
Special case: $a + b + c = 0 \Rightarrow a^{3} + b^{3} + c^{3} = 3 ab c$; Use for evaluation.
Reciprocal sum: $x + \frac{1}{x} = m \Rightarrow x^{3} + \frac{1}{x ^{3}} = m^{3} - 3 m$; And $x - \frac{1}{x} = n \Rightarrow x^{3} - \frac{1}{x ^{3}} = n^{3} + 3 n$ .

How to solve Cubic Identities and the Sum of Three Cubes questions

Decide first whether the task is to expand, factorise or evaluate; then pick the matching cube identity.

1Identify cube terms: write each term as $(\dots)^{3}$ , such as $64 x^{3} = (4 x)^{3}$ and $125 = 5^{3}$ .
2For expansion use the pattern $1, 3, 3, 1$ with falling powers of the first term and rising powers of the second; check the signs.
3For $a^{3} \pm b^{3}$ write the linear factor from the cube roots, then build the quadratic factor with the opposite sign on the middle term.
4For numbers, split into base and small part, such as $103 = 100 + 3$ .
5For values from sum or product data, use the rearranged forms $a^{3} + b^{3} = (a + b)^{3} - 3 ab (a + b)$ .
6For three cubes, test $a + b + c = 0$ first.
7Check with a small number such as $a = 1$ , $b = 1$ .

Cubic Identities and the Sum of Three Cubes in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Cube of a sum

Suppose a toy maker enlarges a wooden cube of edge 4 cm by adding a 1 cm layer on every edge, so the new edge is 5 cm.

By how much does the volume increase, and which parts of $(4 + 1)^{3}$ account for it?

Try it first, then show the solutionSolution

Old volume $4^{3} = 64$ cm $^{3}$ ; new volume $5^{3} = 125$ cm $^{3}$ .
Increase $= 125 - 64 = 61$ cm $^{3}$ .
From $(4 + 1)^{3} = 4^{3} + 3 (4^{2}) (1) + 3 (4) (1^{2}) + 1^{3}$ the extra volume is $48 + 12 + 1 = 61$ .

Answer: The volume increases by 61 cm $^{3}$ : three slabs of 16 cm $^{3}$ , three strips of 4 cm $^{3}$ and one corner cube of 1 cm $^{3}$ .

Memory hook: Three slabs, three strips, one corner.

Tricks to remember Cubic Identities and the Sum of Three Cubes

Short memory aids. Say each one aloud once, then use it on a question the same day.

1, 3, 3, 1
The coefficients of a cube of a binomial are 1, 3, 3, 1; the powers of the first term fall and the second rise.
Use it when: Expanding $(a \pm b)^{3}$ .
SOAP
In $a^{3} \pm b^{3}$ : Same sign, Opposite sign, Always Positive: the linear factor keeps the sign of the original, the middle term of the quadratic takes the opposite sign, and the last term is always positive.
Use it when: Writing the factors of sum and difference of cubes.
Zero sum, triple product
If three numbers add to zero, the sum of their cubes is three times their product.
Use it when: Evaluating cube sums with no calculator.

Cubic Identities and the Sum of Three Cubes: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Expand a cube · 3 marks · Short answerBoard level

Expand $(2 x + 3 y)^{3}$ .

Try it first, then show the solutionSolution

Use $(a + b)^{3} = a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ with $a = 2 x$ , $b = 3 y$ . (½)
$a^{3} = 8 x^{3}$ ; $3 a^{2} b = 3 (4 x^{2}) (3 y) = 36 x^{2} y$ . (1)
$3 a b^{2} = 3 (2 x) (9 y^{2}) = 54 x y^{2}$ ; $b^{3} = 27 y^{3}$ . (1)
Sum: $8 x^{3} + 36 x^{2} y + 54 x y^{2} + 27 y^{3}$ . (½)

Answer: $(2 x + 3 y)^{3} = 8 x^{3} + 36 x^{2} y + 54 x y^{2} + 27 y^{3}$ .

Example 2: Difference of cubes · 3 marks · Short answerBoard level

Factorise $27 a^{3} - 8 b^{3}$ .

Try it first, then show the solutionSolution

$27 a^{3} = (3 a)^{3}$ and $8 b^{3} = (2 b)^{3}$ . (1)
Use $p^{3} - q^{3} = (p - q) (p^{2} + pq + q^{2})$ with $p = 3 a$ , $q = 2 b$ . (1)
$= (3 a - 2 b) (9 a^{2} + 6 ab + 4 b^{2})$ . (1)

Answer: $27 a^{3} - 8 b^{3} = (3 a - 2 b) (9 a^{2} + 6 ab + 4 b^{2})$ .

Example 3: Sum and product data · 3 marks · Short answerHarder board level

If $x + y = 8$ and $x y = 15$ , find $x^{3} + y^{3}$ and $x^{3} - y^{3}$ (both possible values).

Try it first, then show the solutionSolution

$x^{3} + y^{3} = (x + y)^{3} - 3 x y (x + y) = 512 - 3 (15) (8) = 512 - 360 = 152$ . (1)
$(x - y)^{2} = (x + y)^{2} - 4 x y = 64 - 60 = 4$ , so $x - y = \pm 2$ . (1)
$x^{2} + x y + y^{2} = (x + y)^{2} - x y = 64 - 15 = 49$ , so $x^{3} - y^{3} = (x - y) (49) = \pm 98$ . (1)

Answer: $x^{3} + y^{3} = 152$ and $x^{3} - y^{3} = \pm 98$ .

Example 4: Zero-sum cubes · 5 marks · Long answerHarder board level

Prove that if $a + b + c = 0$ then $a^{3} + b^{3} + c^{3} = 3 ab c$ . Hence evaluate $(- 12)^{3} + 7^{3} + 5^{3}$ .

Try it first, then show the solutionSolution

Start from $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (a^{2} + b^{2} + c^{2} - ab - b c - c a)$ . (1)
Prove it by expanding the right side: $a (a^{2} + b^{2} + c^{2} - ab - b c - c a) = a^{3} + a b^{2} + a c^{2} - a^{2} b - ab c - a^{2} c$ and similarly for $b$ and $c$ ; the mixed terms cancel in pairs, leaving $a^{3} + b^{3} + c^{3} - 3 ab c$ . (1)
If $a + b + c = 0$ the right side is $0$ , so $a^{3} + b^{3} + c^{3} = 3 ab c$ . (1)
Here $- 12 + 7 + 5 = 0$ , so the sum of cubes is $3 \times (- 12) \times 7 \times 5$ . (1)
$= - 1260$ . Direct check: $- 1728 + 343 + 125 = - 1260$ . (1)

Answer: The identity is proved and $(- 12)^{3} + 7^{3} + 5^{3} = - 1260$ .

Example 5: Two metal cubes · 4 marks · Case-basedHOTS / competency

Case: two solid metal cubes have edges $p$ cm and $q$ cm. Their edges add up to 11 cm and the sum of their volumes is 737 cm $^{3}$ . (i) Find $pq$ . (ii) Find $p$ and $q$ if $p > q$ . (iii) Find the difference of their volumes. (iv) If the two cubes were melted into one cube, would its edge be a whole number of centimetres?

Try it first, then show the solutionSolution

(i) $p^{3} + q^{3} = (p + q)^{3} - 3 pq (p + q)$ , so $737 = 1331 - 33 pq$ , giving $33 pq = 594$ and $pq = 18$ . (1)
(ii) $p + q = 11$ and $pq = 18$ : $t^{2} - 11 t + 18 = (t - 9) (t - 2) = 0$ , so $p = 9$ and $q = 2$ . Check $729 + 8 = 737$ . (1)
(iii) Difference of volumes $= 729 - 8 = 721$ cm $^{3}$ . (1)
(iv) $9^{3} = 729 < 737 < 1000 = 1 0^{3}$ , so the edge lies between 9 and 10 and is not a whole number. (1)

Answer: (i) $pq = 18$ ; (ii) $p = 9$ , $q = 2$ ; (iii) $721$ cm $^{3}$ ; (iv) no.

Where students lose marks in Cubic Identities and the Sum of Three Cubes

The slips examiners see most often, and what to write instead.

Mistake: Wrong sign in the quadratic factor
Copying the sign from the linear factor.
Wrong working
$a^{3} + b^{3} = (a + b) (a^{2} + ab + b^{2})$ .
Right working
$a^{3} + b^{3} = (a + b) (a^{2} - ab + b^{2})$ .
Do this: Use SOAP: same, opposite, always positive.
Usually costs: 1 to 2 marks.
Mistake: Forgetting the factors 3 in the cube expansion
Writing $a^{3} + a^{2} b + a b^{2} + b^{3}$ .
Wrong working
$(x + 2)^{3} = x^{3} + 2 x^{2} + 4 x + 8$ .
Right working
$(x + 2)^{3} = x^{3} + 6 x^{2} + 12 x + 8$ .
Do this: Write the coefficients 1, 3, 3, 1 before substituting.
Usually costs: Usually 2 of 3 marks.
Mistake: Not squaring the coefficient of the variable
$(2 x)^{2}$ written as $2 x^{2}$ .
Wrong working
$(2 x + 3 y)^{3}$ middle term $3 \cdot 2 x^{2} \cdot 3 y$ .
Right working
$3 (2 x)^{2} (3 y) = 36 x^{2} y$ .
Do this: Put a bracket around each term before powering.
Usually costs: 1 mark.

Exam tips

Check whether the numbers add to zero before cubing anything.
Write each term as a cube, for example $27 a^{3} = (3 a)^{3}$ , before applying the identity.
For reciprocal questions, derive each power step by step as in the worked example; do not memorise a table.
A final numerical check with $x = 1$ or $2$ catches sign errors.

Practice questions on Cubic Identities and the Sum of Three Cubes

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
The value of $9 9^{3}$ is
(a) 970299
(b) 960299
(c) 970199
(d) 980299
Hint
Write $99 = 100 - 1$ and expand.
Answer
(a) 970299. $(100 - 1)^{3} = 1000000 - 30000 + 300 - 1 = 970299$ .
Q2 · 3 marks · Short answerBoard level
Factorise $64 x^{3} + 125$ .
Hint
Both terms are perfect cubes: $(4 x)^{3}$ and $5^{3}$ .
Answer
$(4 x)^{3} + 5^{3} = (4 x + 5) (16 x^{2} - 20 x + 25)$ .
Q3 · 3 marks · Short answerHarder board level
If $x - \frac{1}{x} = 4$ , find $x^{2} + \frac{1}{x ^{2}}$ and $x^{3} - \frac{1}{x ^{3}}$ .
Hint
Square first; then use $n^{3} + 3 n$ for the cube.
Answer
$x^{2} + \frac{1}{x ^{2}} = 4^{2} + 2 = 18$ . $x^{3} - \frac{1}{x ^{3}} = 4^{3} + 3 (4) = 64 + 12 = 76$ ; check with $(x - \frac{1}{x}) (x^{2} + 1 + \frac{1}{x ^{2}}) = 4 \times 19 = 76$ .
Q4 · 5 marks · Long answerHarder board level
Factorise completely $x^{6} - y^{6}$ , and verify your answer at $x = 2$ , $y = 1$ .
Hint
Treat it as $(x^{3})^{2} - (y^{3})^{2}$ first.
Answer
$x^{6} - y^{6} = (x^{3} - y^{3}) (x^{3} + y^{3}) = (x - y) (x^{2} + x y + y^{2}) (x + y) (x^{2} - x y + y^{2})$ . Check: LHS $= 64 - 1 = 63$ ; RHS $= (1) (4 + 2 + 1) (3) (4 - 2 + 1) = 7 \times 3 \times 3 = 63$ .
Q5 · 4 marks · Long answerHOTS / competency
Rhea says "for any three numbers, $a^{3} + b^{3} + c^{3} = 3 ab c$ ". (i) Test her claim for $a = 1$ , $b = 2$ , $c = 3$ . (ii) State the exact condition under which the claim holds. (iii) For $a = 1$ and $b = 2$ , find the value of $c$ that makes the claim true and verify. (iv) Is there any other value of $c$ ?
Hint
Use $a^{3} + b^{3} + c^{3} - 3 ab c = (a + b + c) (\dots)$ and ask when each factor vanishes.
Answer
(i) $1 + 8 + 27 = 36$ but $3 ab c = 18$ , so the claim is false. (ii) It holds when $a + b + c = 0$ or when $a = b = c$ . (iii) $c = - 3$ : $1 + 8 - 27 = - 18$ and $3 (1) (2) (- 3) = - 18$ . (iv) The second factor is zero only when $a = b = c$ , which is impossible with $a = 1$ , $b = 2$ , so $c = - 3$ is the only value.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1. $(a + b)^{3}$ ?Show answer
$a^{3} + 3 a^{2} b + 3 a b^{2} + b^{3}$ .
2. $a^{3} - b^{3}$ ?Show answer
$(a - b) (a^{2} + ab + b^{2})$ .
3.If $a + b + c = 0$ , then $a^{3} + b^{3} + c^{3}$ ?Show answer
$3 ab c$ .
4. $x + \frac{1}{x} = m$ gives $x^{3} + \frac{1}{x ^{3}}$ ?Show answer
$m^{3} - 3 m$ .
5.When is $a^{3} + b^{3} + c^{3} = 3 ab c$ ?Show answer
When $a + b + c = 0$ or $a = b = c$ .

Cubic Identities and the Sum of Three Cubes: frequently asked questions

Does $a^2 - ab + b^2$ factorise further?

Not over the real numbers. Its discriminant is negative, so it has no real linear factors.

How do I remember the signs in the sum and difference of cubes?

Use SOAP. For a^3 + b^3 the linear factor is a + b (same), the middle term of the quadratic is -ab (opposite), and the last term +b^2 is always positive.

Why does $a + b + c = 0$ give $a^3+b^3+c^3 = 3abc$?

Because a^3+b^3+c^3-3abc = (a+b+c)(\ldots), and the first factor is zero.