QPaperGen

Let one number be \( x \).
Then the other number is \( 16 - x \).
We want to minimize the sum of their cubes:
\[
S(x) = x^3 + (16 - x)^3
\]

Differentiate:
\[
S'(x) = 3x^2 - 3(16 - x)^2
\]

Set \( S'(x) = 0 \) for critical points:
\[
3x^2 = 3(16 - x)^2 \Rightarrow x^2 = (16 - x)^2
\]

Expand:
\[
x^2 = 256 - 32x + x^2 \Rightarrow -32x + 256 = 0
\Rightarrow x = \frac{256}{32} = 8
\]

Now check second derivative:
\[
S''(x) = 6x + 6(16 - x) = 6x + 96 - 6x = 96 > 0
\]

Since \( S''(8) > 0 \), this confirms a local minimum.

\[\] Final Answer:
\[
\boxed{\text{The two numbers are } 8 \text{ and } 16-8=8.}
\]