3. Find the real solutions of the equation: $x\left(\frac{6-x}{x+1}\right)\left(\frac{6-x}{x+1}+x\right)=8$.

## SOLUTION:
Let $t=x+1$. Then substituting into the given equation, we have $(t-1)\left(\frac{7}{t}-1\right)\left(\frac{7}{t}+t-2\right)=8$, which is equivalent to $\left(7-t-\frac{7}{t}+1\right)\left(\frac{7}{t}+t-2\right)=8$. Making the substitution $\quad u=\frac{7}{t}+t$, we obtain the equation $(8-u)(u-2)=8$ equivalent to $u^{2}-10 u+24=0$, whose solutions are $u=4$ and $u=6$.

From $u=\frac{7}{t}+t=4$, we get $t^{2}-4 t+7=0$ which has no real solutions and if $u=\frac{7}{t}+t=6$, we arrive at the equation $t^{2}-6 t+7=0$, whose two solutions are real: $t_{1}=3+\sqrt{2}$ and $t_{2}=3-\sqrt{2}$. Since $x=t-1$, the initial equation has the real solutions $x_{1}=2-\sqrt{2}$ and $x_{2}=2+\sqrt{2}$.