Two Questions on Cube Identities (Class 9)

This lesson solves two sure questions on cube identities from polynomials. The first evaluates a difference of cubes from a linear relation and a product; the second proves that a sum of three cubes factorises when the three bases add to zero.

Question 1: evaluate the difference of cubes

We are given

$4x - 5y = 16, \qquad xy = 12,$

and asked to find

$64x^3 - 125y^3.$

First rewrite the target as a difference of cubes:

$64x^3 - 125y^3 = (4x)^3 - (5y)^3.$

The identity

Use the cube identity

$a^3 - b^3 = (a - b)^3 + 3ab(a - b),$

with $a = 4x$ and $b = 5y$. Then

$a - b = 4x - 5y = 16, \qquad ab = (4x)(5y) = 20xy = 20 \times 12 = 240.$

Substituting and computing

$(4x)^3 - (5y)^3 = (16)^3 + 3(240)(16).$

Work out each piece:

$16^3 = 16 \times 256 = 4096, \qquad 3 \times 240 \times 16 = 720 \times 16 = 11520.$

Adding,

$4096 + 11520 = 15616.$

So $64x^3 - 125y^3 = 15616$.

Question 2: sum of three cubes

Without actually expanding the cubes, show that

$(p-q)^3 + (q-r)^3 + (r-p)^3 = 3(p-q)(q-r)(r-p).$

Naming the three terms

Let

$a = p - q, \qquad b = q - r, \qquad c = r - p.$

Then

$a + b + c = (p-q) + (q-r) + (r-p) = 0,$

since every variable cancels.

The key identity

Whenever $a + b + c = 0$, we have

$a^3 + b^3 + c^3 = 3abc.$

Applying this here,

$(p-q)^3 + (q-r)^3 + (r-p)^3 = 3(p-q)(q-r)(r-p).$

That is the required result.

Key takeaways

• A difference of cubes can be written as $a^3 - b^3 = (a-b)^3 + 3ab(a-b)$, which lets you evaluate it from $a-b$ and $ab$ alone.
• Here $a-b = 16$ and $ab = 20xy = 240$ give $64x^3 - 125y^3 = 15616$.
• If $a + b + c = 0$, then $a^3 + b^3 + c^3 = 3abc$, so $(p-q)^3 + (q-r)^3 + (r-p)^3 = 3(p-q)(q-r)(r-p)$.