<s>[INST] Hints on solving $x^{2x}-(x^2+x)x^x+x^3=0$ 
Solve this equation over $\mathbb{R}^+$: $x^{2x}-(x^2+x)x^x+x^3=0$

I’ve been trying to solve this exponential equation but can’t get the answer because normal substitution ($y=x^x$) isn’t working. Any tips/hints that don’t use logs?
(the section of the book I got this from is before the introduction of logarithms). Thanks.
 [/INST] There are only three obvious solutions
$x=-1,x=1$ and $x=2$
The other intervals are inconsistent with the sign of the simplified equation:
$x^{2x}-(x^2+x)x^x+x^3$ = 0  { expand }
$x^3+x^{2x}-x^{1+x}-x^{2+x}=0$   {a}
$1+x^{-3+2x}-x^{-2+x}-x^{-1+x}=0$  Divide {a} by $x^3$
Simplified to
$-x^{x-2}-x^{x-1}+x^{x^2-3}=-1$   (1)
for example if x>2
$-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive sign where is inconsistent with the equation simplified (1):
Cause: when x>2
$-x^{x-2}-x^{x-1} < x^{x^2-3}$ So
$-x^{x-2}-x^{x-1}+x^{x^2-3}$ is always positive.
So No solution for x>2.
You can continued the demonstration with the intervals : 1<x<2 , 0<x<1, x=0 , -1<x<0 , x<-1 . All  this intervals gives no solutions in Reals.
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