6.171. $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$

6.171. $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$
## Solution.

After raising both sides of the equation to the third power, we have

$$
x+3 \sqrt[3]{x^{2}(x-16)}-3 \sqrt[3]{x(x-16)^{2}}+x-16=x-8 \Leftrightarrow
$$

$$
\Leftrightarrow 3 \sqrt[3]{x(x-16)}(\sqrt[3]{x}+\sqrt[3]{x-16})=8-x
$$

Since by the condition $\sqrt[3]{x}+\sqrt[3]{x-16}=\sqrt[3]{x-8}$, we have

$$
3 \sqrt[3]{x(x-16)(x-8)}=8-x
$$

From this,

$$
\begin{aligned}
& 27 x(x-16)(x-8)=(8-x)^{3} \Leftrightarrow 27 x(x-16)(x-8)+(x-8)^{3}=0 \\
& (x-8)\left(27 x(x-16)+(x-8)^{2}\right)=0,(x-8)\left(7 x^{2}-112 x+16\right)=0 \Rightarrow x-8=0
\end{aligned}
$$

$x_{1}=8$, or $7 x^{2}-112 x+16=0$, from which $x_{2,3}=\frac{56 \pm \sqrt{3024}}{7}$.

Answer: $x_{1}=8, x_{2,3}=8 \pm \frac{12 \sqrt{21}}{7}$.