How to solve for $x$ in $x^3+8^x-9=0$

Question

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

Answer 1

$$(x^3 -1)=(8-8^x) $$

$$(x-1)\color{Violet}{(x^2 +x+1)}= (8-8^x )$$

$$But\color{Violet} {(x^2+x+1)>0}$$

$$Eithr\color{blue}{ (x-1)\geq0} \space\space and\space \color{red}{8-8^x\geq0}$$

$$\color{blue}{x\geq1}\space and \space \color{red}{x\leq1} \Rightarrow\space \color{green} {x=1}$$

$$OR\color{blue}{ (x-1)\leq0} \space\space and\space \color{red}{8-8^x\leq0}$$

$$\color{blue}{x\leq1}\space and \space \color{red}{x\geq1} \Rightarrow\space \color{green} {x=1}$$

Answer 2

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.

Answer 3

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.