How to solve for $x$ in $x^3+8^x-9=0$ How to find real numbers $x$ that are solutions to
$$x^3+8^x-9=0$$
Please help me.

Note that
\begin{align}
x^3+8^x-9=0\iff
&
\begin{cases}
x^3-1=2^3-(2^{x})^3\\
x^3-8=1^3-(2^{x})^3\\
\end{cases}
\\
\iff 
&
\begin{cases}
(x-1)(x^2+x+1)=[2-(2^{x})][2^2+2(2^{x}) + (2^{x})^2] \\
(x-2)(x^2+2x+4)=[1-(2^{x})][1+(2^{x})+(2^{x})^2] \\
\end{cases}
\\
\end{align}
Then is easy to see that $x=1 \implies x^3+8^x-9=0 $.
We have that $f(x)=x^3+8^x-9\implies f^\prime(x)=3x^2+e^x\cdot\log_e(8)>0$ and then $f(x)$ have only one root. That is, $f(x)=0\implies x=1$. 
Then $x=1$ the unique root.
