Product of Squares from Sum of Cubes

$a^3 + b^3 = 128$ $\frac{1}{a^3} + \frac{1}{b^3} = \frac{a^3 + b^3}{a^3 b^3} = 2 \implies \frac{128}{a^3 b^3} = 2 \implies a^3 b^3 = 64 \implies ab = 4$. Product of squares $= (ab)^2 = 4^2 = 16$.