Let

x and y = the numbers

z = sum of their cubes

$k = x + y$

$y = k - x$

$z = x^3 + y^3$

$z = x^3 + (k - x)^3$

$dz/dx = 3x^2 + 3(k - x)^2(-1) = 0$

$x^2 - (k^2 - 2kx + x^2) = 0$

$x = \frac{1}{2}k$

$y = k - \frac{1}{2}k$

$y = \frac{1}{2}k$

$z = (\frac{1}{2}k)^3 + (\frac{1}{2}k)^3$

$z = \frac{1}{4}k^3$