Apéry's constant by triple integral I would like to ask if someone would help me solve the following integral.
\begin{equation}
   \zeta(3) = \int_0^1 \int_0^1 \int_0^1 \frac{1}{1-xyz}\ dx\,dy\,dz
\end{equation}
It occurred to me that if I transformed the integral into spherical, cylindrical, polar coordinates, or turned the volume created by integrals. Maybe it would help in solving.
Thank you in advance.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
\begin{align}
&\bbox[10px,#ffd]{\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
{\dd x\,\dd y\,\dd z \over 1 - xyz}} =
\int_{0}^{1}\int_{0}^{1}\int_{0}^{1}
\bracks{\sum_{n = 0}^{\infty}\pars{xyz}^{n}}\dd x\,\dd y\,\dd z
\\[5mm] = &\
\sum_{n = 0}^{\infty}\pars{\int_{0}^{1}x^{n}\,\dd x}
\pars{\int_{0}^{1}y^{n}\,\dd y}
\pars{\int_{0}^{1}z^{n}\,\dd z} =
\sum_{n = 0}^{\infty}{1 \over \pars{n + 1}^{3}}
\\[5mm] = &\
\sum_{n = 1}^{\infty}{1 \over n^{3}} = \bbx{\zeta\pars{3}}
\end{align}
