Solve the equation $8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$ Solve the equation $$8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}$$ The given equation is equivalent to $$2^{3x}+\dfrac{12}{2^x}=1+\dfrac{8}{2^{3x}}+6\cdot2^x$$ If we put $a:=2^x>0$, the equation becomes $$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$ which is $$a^6-6a^4-a^3+12a^2-8=0$$ The LHS factors as $(a+1)(a-2)(a^4+a^3-3a^2-2a+4)$, which is in no case obvious. Let's say that we find the roots $1$ and $-2$, then how do we show that $(a^4+a^3-3a^2-2a+4)$ does not factor any more? Taking these into consideration, I believe there is an another approach. Any ideas would be appreciated.

Letting $y=2^{x-\frac12}>0$
$$\begin{align}8^x+3\cdot2^{2-x}=1+2^{3-3x}+3\cdot2^{x+1}&\iff y^3+\frac3y=\frac1{2\sqrt2}+\frac1{y^3}+3y
\\&\iff\left(y-\frac1y\right)^3=\left(\frac1{\sqrt2}\right)^3\\&\iff\dots\\&\iff y=\sqrt2\\&\iff x=1.\end{align}$$
