Giải thích các bước giải:
đặt: \[\begin{array}{l}
P = \frac{{(6 - 2x)(12 - 3y)(2x - 3y)}}{6} \le \frac{{{{(6 - 2x + 12 - 3y + 2x + 3y)}^3}}}{{6.27}} = \frac{{{{18}^3}}}{{6.27}} = 36\\
\left\{ {\begin{array}{*{20}{c}}
{x = \frac{a}{{{b^3}}}}\\
{y = \frac{b}{{{c^3}}}}\\
{z = \frac{c}{{{a^3}}}}
\end{array} \Rightarrow \left\{ {\begin{array}{*{20}{c}}
{\frac{1}{x} = \frac{{{b^3}}}{a}}\\
{\frac{1}{y} = \frac{{{c^3}}}{b}}\\
{\frac{1}{z} = \frac{{{a^3}}}{c}}
\end{array}} \right.} \right. \Rightarrow xyz = 1
\end{array}\]
\[\begin{array}{l}
\Rightarrow x + y + z = \frac{1}{x} + \frac{1}{y} + \frac{1}{z}\\
\Rightarrow x + y + z = yz + xz + xy\\
\Rightarrow xyz + (x + y + z) - (xy + yz + zx) - 1 = 0\\
\Rightarrow (x - 1)(y - 1)(z - 1) = 0
\end{array}\]
vậy tồn tại x=1=> \[a = {b^3}\](dpcm)
