Question 74
Find all positive reals $x, y, z$ so that
$$
\begin{cases}
xyz = 6 \\
(x+1)(y+2)(z+3) = 48.
\end{cases}
$$
$$
\begin{cases}
xyz = 6 \\
(x+1)(y+2)(z+3) = 48.
\end{cases}
$$
MathNet 2016-part-04 — 184 problems (EN–VI) Question 74
Math Olympiad Vault · MathNet 2016-part-04 — 184 problems (EN–VI)