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Choose the factorised form of (64m^3+125n^3).

Answer and explanation

Correct answer: (4m+5n)(16m^2-20mn+25n^2)

The direct answer is A: $(4m+5n)(16m^2-20mn+25n^2)$. Recognize $64m^3=(4m)^3$ and $125n^3=(5n)^3$. Use the sum-of-cubes identity $u^3+v^3=(u+v)(u^2-uv+v^2)$, with $u=4m$ and $v=5n$. Substitution gives $(4m+5n)[(4m)^2-(4m)(5n)+(5n)^2]=(4m+5n)(16m^2-20mn+25n^2)$. Option A is correct. B uses $4m-5n$ in the first factor, which belongs to a difference pattern, not this sum. C has a positive middle term; multiplying it creates the wrong mixed terms. D incorrectly uses $64m+125n$ instead of the cube roots $4m$ and $5n$, and its second factor is not the correct identity factor. A useful memory cue is: for a sum of cubes, signs are plus, minus, plus in the two factors: $(u+v)(u^2-uv+v^2)$.

Related tags

Sum Of CubesCubic IdentitiesFactorisationExploring Algebraic IdentitiesMathematicsClass 9 Mcq

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