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What is the factorised form of (27a^3+64b^3)?

Answer and explanation

Correct answer: ((3a+4b)(9a^2-12ab+16b^2))

The direct answer is A: \(27a^3+64b^3=(3a+4b)(9a^2-12ab+16b^2)\). Use the sum-of-cubes identity \(p^3+q^3=(p+q)(p^2-pq+q^2)\). Put \(p=3a\) and \(q=4b\). Then \(p^3=27a^3\), \(q^3=64b^3\), \(p+q=3a+4b\), and the second factor becomes \(9a^2-12ab+16b^2\). Option A is correct. Option B uses the difference-of-cubes first factor \(3a-4b\), so it does not represent a sum. Option C has an incorrect structure and does not follow the cube identity. Option D has the wrong sign in the middle term: for a sum of cubes, the second factor has a negative middle term, not a positive one. Multiplying A confirms the original expression because the cross terms cancel appropriately. Memory cue: sum of cubes has plus, then plus-minus-plus; difference has minus, then all plus.

Related tags

Sum Of CubesFactorisationIdentity

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