<s>[INST] Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$ Find all real solutions for $x$ in $2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 .$
I have found out that the answers were 0,1, and -1. But I used sort of a guess-and check way.
$2(2^x-1)x^2+(2^{x^2}-2)x=2^{x+1}-2$
I expanded it into:
$(x^2-1)(2^{x+1}-2)+x(2^{x^2}-2)=0$
I just found all the possibilities to make each group of equation 0 resulting in 0,1, and -1. 
How can I prove this is correct instead of trial/error?
 [/INST] Mistake when you divide by 2. You get:
 $$2^{x^2-1} = 2^{(x+1)(x-1)} = 2^{{x+1}^{x-1}}$$
Now maybe it gets easier?
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