6.139. $(x+1)^{5}+(x-1)^{5}=32 x$.

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6.139. $(x+1)^{5}+(x-1)^{5}=32 x$.
## Solution.

Since $a^{n}+b^{n}=(a+b)\left(a^{n-1}-a^{n-2} b+a^{n-3} b^{2}-a^{n-4} b^{3}+\ldots+b^{n-1}\right)$, then $(x+1+x-1)\left((x+1)^{4}-(x+1)^{3}(x-1)+(x+1)^{2}(x-1)^{2}-(x+1)(x-1)^{3}+\right.$ $\left.+(x-1)^{4}\right)=32 x, \quad 2 x\left((x+1)^{4}+(x-1)^{4}-(x+1)^{3}(x-1)-(x+1)(x-1)^{3}+\right.$ $\left.+(x+1)^{2}(x-1)^{2}\right)-32 x=0 ; \quad x\left((x+1)^{4}+(x-1)^{4}-(x+1)^{3}(x-1)-(x+1) \times\right.$ $\left.\times(x-1)^{3}+(x+1)^{2}(x-1)^{2}-16\right)=0 \Rightarrow \quad x_{1}=0, \quad$ or $(x+1)^{4}+(x-1)^{4}-$ $-(x+1)^{3}(x-1)-(x+1)(x-1)^{3}+(x+1)^{2}(x-1)^{2}-16=0$.

Since $a^{4}+b^{4}=\left((a-b)^{2}+2 a b\right)^{2}-2 a^{2} b^{2}$, we have:

$$
\left((x+1-x+1)^{2}+2(x+1)(x-1)\right)^{2}-2(x+1)^{2}(x-1)^{2}-(x+1)(x-1) \times
$$

$$
\begin{aligned}
& \times\left((x+1)^{2}+(x-1)^{2}\right)+(x+1)^{2}(x-1)^{2}-16=0, \\
& -(x+1)^{2}(x-1)^{2}-(x+1)(x-1)\left((x+1)^{2}+(x-1)^{2}\right)-16=0,(4+2(x+1) \times \\
& \times(x-1))^{2}-(x+1)^{2}(x-1)^{2}-(x+1)(x-1)\left(\left(x+1-x+1^{2}\right)+2(x+1)(x-1)\right)- \\
& -16=0, \text { or }\left(4+2\left(x^{2}-1\right)\right)^{2}-\left(x^{2}-1\right)^{2}-\left(x^{2}-1\right)\left(4+2\left(x^{2}-1\right)\right)-16=0 .
\end{aligned}
$$

Let $x^{2}-1=y$. We have $(4+2 y)^{2}-y^{2}-y(4+2 y)-16=0$, $16+16 y+4 y^{2}-y^{2}-4 y-2 y^{2}-16=0, \quad y^{2}+12 y=0, \quad y(y+12)=0 \Rightarrow$ $\Rightarrow y_{1}=0, y_{2}=-12$. Then $x^{2}-1=0, x_{2,3}= \pm 1$, or $x^{2}-1=-12$, from which $x^{2}=-11<0$, which is not valid.

Answer: $x_{1}=0, x_{2}=-1, x_{3}=1$.