Which expression can be factorised directly using the identity for the sum of cubes, \(a^3+b^3=(a+b)(a^2-ab+b^2)\)?

Correct answer: \(125x^3+64y^3\)

The correct answer is A: \(125x^3+64y^3\). A sum of cubes must have two positive cube terms. Here \(125x^3=(5x)^3\) and \(64y^3=(4y)^3\), so option A is exactly \((5x)^3+(4y)^3\). It can be factorised as \((5x+4y)(25x^2-20xy+16y^2)\) using \(a^3+b^3=(a+b)(a^2-ab+b^2)\). Option B is a difference of cubes, not a sum, so it requires the difference-of-cubes identity. Option C contains squares rather than cubes, so the stated identity does not apply. Option D has \(16y^3\), but 16 is not a perfect cube, so it cannot be written as \((\text{integer times }y)^3\) in the required direct form. The memory cue is to inspect both the exponent and the numerical coefficient: both terms must be perfect cubes with a plus sign.