Ta có:
\(\dfrac{x^3}{8}=\dfrac{y^3}{64}=\dfrac{z^3}{216}\Rightarrow\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}\)
Đặt \(\dfrac{x}{2}=\dfrac{y}{4}=\dfrac{z}{6}=k\Rightarrow\left\{{}\begin{matrix}x=2k\\y=4k\\z=6k\end{matrix}\right.\)
Mà x2 + y2 + z2 = 14
=> (2k)2 + (4k)2 + (6k)2 = 14
=> 4k2 + 16k2 + 36k2 = 14
=> (4 + 16 + 36)k2 = 14
=> 56k2 = 14
\(\Rightarrow k^2=\dfrac{14}{56}=\dfrac{1}{4}\)
\(\Rightarrow k=\pm\dfrac{1}{2}\)
- Với \(k=\dfrac{1}{2}\) thì ta có:
\(x=2\cdot\dfrac{1}{2}=1\)
\(y=4\cdot\dfrac{1}{2}=2\)
\(z=6\cdot\dfrac{1}{2}=3\)
- Với \(k=-\dfrac{1}{2}\) thì ta có:
\(x=2\cdot\left(-\dfrac{1}{2}\right)=-1\)
\(y=4\cdot\left(-\dfrac{1}{2}\right)=-2\)
\(z=6\cdot\left(-\dfrac{1}{2}\right)=-3\)
Vậy x = 1, y = 2, z = 3 hoặc x = -1, y = -2, z = -3
