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.
 A: HINT: Try a few small whole numbers, and you’ll quickly find a solution. Note also that $x^3+8^x$ is an increasing function of $x$, so there can be at most one solution.
A: $$(x^3 -1)=(8-8^x) $$
$$(x-1)\color{Violet}{(x^2 +x+1)}= (8-8^x )$$
$$But\color{Violet} {(x^2+x+1)>0}$$
$$Eithr\color{blue}{  (x-1)\geq0} \space\space and\space \color{red}{8-8^x\geq0}$$
$$\color{blue}{x\geq1}\space and \space \color{red}{x\leq1} \Rightarrow\space \color{green} {x=1}$$
$$OR\color{blue}{  (x-1)\leq0} \space\space and\space \color{red}{8-8^x\leq0}$$
$$\color{blue}{x\leq1}\space and \space \color{red}{x\geq1} \Rightarrow\space \color{green} {x=1}$$
A: 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.
