Let
x and y = the numbers
z = sum of their cubes
$k = x + y$
$y = k - x$
$z = x^3 + y^3$
$z = x^3 + (k - x)^3$
$dz/dx = 3x^2 + 3(k - x)^2(-1) = 0$
$x^2 - (k^2 - 2kx + x^2) = 0$
$x = \frac{1}{2}k$
$y = k - \frac{1}{2}k$
$y = \frac{1}{2}k$
$z = (\frac{1}{2}k)^3 + (\frac{1}{2}k)^3$
$z = \frac{1}{4}k^3$