<s>[INST] Find roots of $3^x+x^3=17$ Find $x$ in the following equation:
$$3^x+x^3=17$$
 [/INST] Note that if $y\gt 0$we have $3^y\gt 1$ so that if $x\ge 0$ $$3^{x+y}+(x+y)^3=3^x\cdot 3^y +x^3+(3x^2y+3xy^2+y^3)\gt 3^x+x^3$$
And it is very easy to show that for $x\le 0$ we have $3^x+x^3\le 1+0\lt 17$
So the function is increasing for positive $x$ (without calculus) and there is at most a single solution.
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