Find all $x$ such that $2^x,2^{x^2}$ and $2^{x^3}$ form $3$ terms of an A.P. I know that if $a,b,c$ are in Arithmetic Progression, then $2b=a+c$, but in this case, I am unable to solve for $x$. Hints are appreciated. Thanks.
HINT:

$$2^x(1+2^{x^3-x})=2^{x^2+1}\iff 1+2^{x^3-x}=2^{x^2-x+1}$$

If $x^3-x&gt;0,1+2^{x^3-x}&gt;1$ is odd unlike $2^{x^2-x+1}$ as $x^2-x+1&gt;0$ for real $x$

So, $x^3-x$ must be $0$

and consequently, $2=2^{x^2-x+1}\implies 1=x^2-x+1$
Since $2b=a+c$ we have 
$2. 2^{x^{2}}=2^{x}+2^{x^{3}}$ 

which implies that $2^{x^{2}+1}=2^{x}+2^{x^{3}}$

Now taking log on both sides we get
$x^{2}+1$ log2=log $(2^{x}+2^{x^{3}})$ solving this we can find $x$