117. Solve the equation:

a) $(x+a)(x+2a)(x+3a)(x+4a)=3a^4$;

b) $x^2+\frac{4}{x^2}+6\left(x+\frac{2}{x}\right)=23$;

c) $\sqrt[5]{\frac{16x}{x-1}}+\sqrt[5]{\frac{x-1}{16x}}=\frac{5}{2}$.
Solution. a) Group the first and fourth, second and third factors of the left side of the equation. We will have:

$$
\left(x^{2}+5 a x+4 a^{2}\right)\left(x^{2}+5 a x+6 a^{2}\right)=3 a^{4}
$$

If $x^{2}+5 a x=y$, then we have $\left(y+4 a^{2}\right)\left(y+6 a^{2}\right)=3 a^{4}$, from which $y^{2}+$ $+10 a^{2} y+21 a^{4}=0, y_{1}=-3 a^{2}, y_{2}=-7 a^{2}$.

To determine $x$, we have two quadratic equations $x^{2}+5 a x+$ $+3 a^{2}=0$ and $x^{2}+5 a x+7 a^{2}=0$, the second of which has no real roots.

b) If $x+\frac{2}{x}=y$, then $x^{2}+\frac{4}{x^{2}}=y^{2}-4$. We have the equation $y^{2}+$ $+6 y-27=0$, from which $y_{1}=3, y_{2}=-9$. Therefore, $x+\frac{2}{x}=3$, $x+\frac{2}{x}=-9$.

Solving the quadratic equations $x^{2}-3 x+2=0$ and $x^{2}+9 x+2=0$, we will find the values of $x$.

c) Let $y=\sqrt[5]{\frac{16 x}{x-1}}$. Then we have the equation $y+\frac{1}{y}=\frac{5}{2}$, or $2 y^{2}-5 y+2=0$, from which $y_{1}=\frac{1}{2}, y_{2}=2$. Therefore,

$$
\begin{gathered}
\frac{16 x}{x-1}=\frac{1}{32}, x_{1}=-\frac{1}{511} \\
\frac{16 x}{x-1}=32, x_{2}=2
\end{gathered}
$$

Answer: a) $\frac{-5 a+a \sqrt{13}}{2} ; \frac{-5 a-a \sqrt{13}}{2} ;$ b) $1 ; 2 ; \frac{-9+\sqrt{73}}{2}$; $\frac{-9-\sqrt{73}}{2} ;$ c) $-\frac{1}{511} ; 2$.