Let x and y be two positive numbers.
$\therefore $ According to question,
$x+y=a \quad$ (constant)
$\Rightarrow y=a-x$
Let $z = x^{3} + y^{3}$
$z = x^{3} + \left(a - x\right)^{3}\quad ... \left(i\right)$
On differentiating Eq. (i) w.r .t. 'x', we get
$\frac{dz}{dx} = 3x^{2}+3\left(a-x\right)^{2} \left(-1\right)$
$= 3\left[x^{2} -\left(a- x\right)^{2}\right]$
$= 3 \left(x + a - x\right) \left(x - a+ x\right)$
$= 3a \left(2x - a\right)\quad ... \left(ii\right)$
For maximum or minimum, $\frac{dz}{dx} = 0$
$\Rightarrow 3a\left(2x- a\right)= 0\quad\left(\because 3a\ne0\right)$
$\Rightarrow 2x-a=0$
$\Rightarrow x = \frac{a}{2}$
On differentiating Eq. (ii) w.r.t 'x', we get
$\frac{d^{2}z}{dx^{2}} = 6a \quad$ (positive)
$\therefore \frac{d^{2}z}{dx^{2}}>0$
$\therefore $ z has a minimum value at $x = \frac{a}{2}$.
Therefore, the numbers are $\frac{a}{2}$ and $a-\frac{a}{2} = \frac{a}{2}.$