If the sides of the rectangle are (fq-8)dm and (f^2g^2 + 8fg + 64) dm. what is the area of the rectangle ?

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2 answers

(fg-8)(f^2g^2 + 8fg + 64)
This is just the difference of two cubes:
(fg)^3 - 8^3

Recall that
(a^3-b^3) = (a-b)(a^2+ab+b^2)
You have
a = fg
b = 8
a = fg+b = 8
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