Factorise 8a³ + 27b³ using sum of cubes

Question

Factorise: $8a^3 + 27b^3$

Solution — Step by Step

The expression $8a^3 + 27b^3$ looks like $x^3 + y^3$ — but $8a^3$ is a cube (since $8a^3 = (2a)^3$ ) and $27b^3$ is a cube (since $27b^3 = (3b)^3$ ).

So we have:

8a^3 + 27b^3 = (2a)^3 + (3b)^3

This is now in the form $x^3 + y^3$ where $x = 2a$ and $y = 3b$ .

The standard identity is:

x^3 + y^3 = (x + y)(x^2 - xy + y^2)

Substituting $x = 2a$ and $y = 3b$ :

8a^3 + 27b^3 = (2a + 3b)\left[(2a)^2 - (2a)(3b) + (3b)^2\right]

= (2a + 3b)\left[4a^2 - 6ab + 9b^2\right]

Each term in the second factor:

$(2a)^2 = 4a^2$
$(2a)(3b) = 6ab$
$(3b)^2 = 9b^2$

\boxed{8a^3 + 27b^3 = (2a + 3b)(4a^2 - 6ab + 9b^2)}

Verification: Expand $(2a + 3b)(4a^2 - 6ab + 9b^2)$ :

$= 2a(4a^2) + 2a(-6ab) + 2a(9b^2) + 3b(4a^2) + 3b(-6ab) + 3b(9b^2)$

$= 8a^3 - 12a^2b + 18ab^2 + 12a^2b - 18ab^2 + 27b^3$

$= 8a^3 + 27b^3$ ✓ (middle terms cancel)

Why This Works

The identity $x^3 + y^3 = (x+y)(x^2 - xy + y^2)$ comes from polynomial division: if you divide $x^3 + y^3$ by $(x+y)$ , you get $x^2 - xy + y^2$ exactly.

The middle term $-xy$ in the second factor is what causes the cancellation of middle terms during expansion. This is why the second factor can’t be further factorised over real numbers — the discriminant of $4a^2 - 6ab + 9b^2$ as a quadratic in $a$ gives $36b^2 - 144b^2 = -108b^2 < 0$ .

Alternative Method — Difference of Cubes

The companion identity: $x^3 - y^3 = (x-y)(x^2 + xy + y^2)$

Note the sign pattern:

Sum of cubes: $(x \mathbf{+} y)(x^2 \mathbf{-} xy + y^2)$ — opposite signs
Difference of cubes: $(x \mathbf{-} y)(x^2 \mathbf{+} xy + y^2)$ — opposite signs again

Memory aid: SOAP — Same, Opposite, Always Positive.

Common Mistake

Writing the second factor as $(4a^2 + 6ab + 9b^2)$ instead of $(4a^2 - 6ab + 9b^2)$ . The middle term MUST be negative for sum of cubes. If you use a positive middle term, expanding will give $8a^3 + 24a^2b + 18ab^2 + 12a^2b + 18ab^2 + 27b^3 \neq 8a^3 + 27b^3$ . A quick expansion check always catches this error.

Question

Solution — Step by Step

Why This Works

Alternative Method — Difference of Cubes

Common Mistake

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