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Question 4

Given three cubes with integer side lengths, if the sum of the surface areas of the three cubes is $$498\text{ cm}^2$$, then the sum of the volumes of the cubes in all possible solutions is

Solution

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If the integer side lengths are $$a,b,c$$, then $$6(a^2+b^2+c^2)=498$$, so $$a^2+b^2+c^2=83$$. The only positive integer solution, up to order, is $$1^2+1^2+9^2=83$$. Therefore the sum of the volumes is $$1^3+1^3+9^3=731$$.

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