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.

Hint
Let's start from
$$a^3+\dfrac{12}{a}=1+\dfrac{8}{a^3}+6a$$
Now rewrite it as
$$\left[a^3-\left(\frac2a\right)^3\right]-6\left[a-\frac 2a\right]-1=0$$
$$\left(a-\frac2a\right)\left[a^2+\frac{4}{a^2}-4\right]-1=0$$
$$\left(a-\frac2a\right)^3-1=0$$
Can you finish?
