<s>[INST] Exponential Equation with strange terms Solve for all $x\in\mathbb{R}$
$(x^4-x^3-x+1)5^{x^2-1}+(x^3-x^2-x+1)5^{x^3-1}+(x^5-x^3-x^2+1)5^{x-1}=x^5+x^4-x^3-2x^2-2x+3$
I rewrote is as $(x^3-1)(x-1)5^{x^2-1}+(x^2-1)(x-1)5^{x^3-1}+(x^2-1)(x^3-1)5^{x-1}=(x^3-1)(x-1)+(x^2-1)(x-1)+(x^3-1)(x^2-1)$
This is equivalent with $(x^3-1)(x-1)(x^2-1)({5^{x^3-1}\over x^3-1}-{1\over x^3-1}+{5^{x^2-1}\over x^2-1}-{1\over x^2-1}+{5^{x-1}\over x-1}-{1\over x-1})=0$
I tried considerind the function $f(x)={5^{x^3-1}\over x^3-1}-{1\over x^3-1}+{5^{x^2-1}\over x^2-1}-{1\over x^2-1}+{5^{x-1}\over x-1}-{1\over x-1}$ which is increasing on $(1,\infty)$(and the function has always positive values) and so x=1 is the only positive solution.
But i don't think this is correct and also im not sure what to do on $(-\infty,1)$ cause -1 is also a solution .
Please help me out it looks kind of ugly and i don't know what to do.
 [/INST] We obtain:
$$(x-1)^2(x^2+x+1)5^{x^2-1}+(x-1)^2(x+1)5^{x^3-1}+(x-1)^2(x+1)(x^2+x+1)5^{x-1}=$$
$$=(x-1)^2(x^3+3x^2+4x+3),$$
which gives $x=1$ or
$$(x^2+x+1)5^{x^2-1}+(x+1)5^{x^3-1}+(x+1)(x^2+x+1)5^{x-1}=x^3+3x^2+4x+3$$ or
$$(x^2+x+1)\left(5^{x^2-1}-1\right)+(x+1)\left(5^{x^3-1}-1\right)+(x+1)(x^2+x+1)\left(5^{x-1}-1\right)=0.$$
Easy to see that $-1$ is a root.
Let $x\neq-1.$
Thus, we need to solve  $f(x)=0,$ where $$f(x)=\frac{5^{x^2-1}-1}{x+1}+\frac{5^{x^3-1}-1}{x^2+x+1}+5^{x-1}-1.$$
Now, prove that $f(1)=0,$ $f(x)>0$ for $x>1$ and $f(x)<0$ for $x<1,$ $x\neq-1$.
</s>