Sum & Difference of Cubes

Unlike a sum of squares, a sum of cubes *does* factor.

a³ + b³ = (a + b)(a² − ab + b²)
a³ − b³ = (a − b)(a² + ab + b²)

The binomial keeps the original sign. In the trinomial, the middle sign is the opposite, and the last is always plus. A common mnemonic is SOAP: Same, Opposite, Always Positive.

For x³ + 8: a = x and b = 2, so it factors as (x + 2)(x² − 2x + 4).

The trinomial almost never factors further, so check whether it does but do not force it. Its discriminant is negative unless a = b.

The identities a³ + b³ = (a + b)(a² − ab + b²) and a³ − b³ = (a − b)(a² + ab + b²) are verified by expansion, where the cross terms cancel.

The quadratic factor is irreducible over ℝ whenever a ≠ b, since its discriminant is b² − 4b² = −3b² (treating a as the variable), which is negative. So the factorisation over ℝ stops there, while over ℂ the quadratic splits into a conjugate pair.

These are the n = 3 cases of the general identity aⁿ − bⁿ = (a − b)(a^{n−1} + a^{n−2}b + … + b^{n−1}), which holds for all n; the sum version requires odd n.

Recognition matters more than memorisation: check whether both terms are perfect cubes (1, 8, 27, 64, 125, 216 and variables with exponents divisible by 3). As always, extract a GCF first — 2x³ − 16 becomes 2(x³ − 8) before the pattern is visible.