$\begin{array}{1 1} 8\;and\;8 \\ 9 \;and\;7 \\ 10\;and\;6 \\ 11 \;and \;5 \end{array} $

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- $\large\frac{d}{dx}$$(x.y)=x.\large\frac{d}{dx}$$(y)+y.\large\frac{d}{dx}$$(x)$

Given

Sum of the No. = 16

Let sum of cubes be S

Let two numbers be $x$ and $16-x$

Sum of cubes $S=x^3+(16-x)^3$

Differentiating with respect to x we get

$\large\frac{dS}{dx}$$=3x^2+3(16-x)^2.(-1)$

$\quad\;\;=3x^2-3(16-x)^2.(-1)$

$\quad\;\;=3(32x-256)$

Step 2:

For minimum value

$\large\frac{dS}{dx}$$=0$

$3(32x-256)=0$

$x=\large\frac{256}{32}$$=8$

Also $\large\frac{d^2S}{dx^2}$$=96>0$

$\Rightarrow S$ is minimum at $x=8$

Hence the required numbers are $8$ & $(16-8)\Rightarrow$ 8 & 8

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