If \(p = 0.8\) and \(q = 0.02\), find the value of \(\dfrac{(4p)^3 + (4q)^3}{p^3 + q^3}\).
64
Pull the constant 4 out of the cubes:
\((4p)^3 + (4q)^3 = 4^3 p^3 + 4^3 q^3 = 64\,(p^3 + q^3)\).
Therefore
\(\dfrac{(4p)^3 + (4q)^3}{p^3 + q^3} = \dfrac{64\,(p^3 + q^3)}{p^3 + q^3} = 64\).
(The specific values of \(p\) and \(q\) are irrelevant since the ratio simplifies to a constant.)
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