58. If positive integers $x, y$ satisfy $x^{3}+3 x^{2} y+8 x y^{2}+6 y^{3}=87$, then $x+2 y=$
Answer: 5.
Solution: Since when $y \geqslant 3$, $6 y^{3}>87$,
at this time
$$
x^{3}+3 x^{2} y+8 x y^{2}+6 y^{3}>87 \text {, }
$$

thus
$y$ can only be 1 or 2.
(1) When $y=1$, substituting gives
$$
\begin{array}{l}
x^{3}+3 x^{2}+8 x+6=87, \\
x^{2}(x+3)+8 x=81,
\end{array}
$$

rearranging, we get

since
$x^{2}(x+3), 8 x$ are both even numbers,

thus
$$
x^{2}(x+3)+8 x \neq 81 \text {. }
$$
(2) When $y=2$, substituting gives

that is
$$
\begin{array}{l}
x^{3}+6 x^{2}+32 x+48=87, \\
x^{3}+6 x^{2}+32 x=39 .
\end{array}
$$

since
$x$ may take the value of 1.
When $x=1$, the equation holds,

thus
$$
x=1, \quad y=2, \quad x+2 y=5 \text {. }
$$