HMMT February

Algebra Difficulty 4.6 AIME Find the answer

Two reals $x$ and $y$ are such that $x-y=4$ and $x^{3}-y^{3}=28$ . Compute $x y$ .

Your answer

A number or a short expression. Fractions can be typed as 3/2, and spacing doesn't matter.

Solution

We have $28=x^{3}-y^{3}=(x-y)\left(x^{2}+x y+y^{2}\right)=(x-y)\left((x-y)^{2}+3 x y\right)=4 \cdot(16+3 x y)$ , from which $x y=-3$ .

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