<s>[INST] Finding $x$ when $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1$ Find $x$ if $\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1.$

I was thinking of trying to substitute some number $y$ written in terms of $x$ than solving for $y$ to solve for $x.$ However, I'm not sure what $y$ to input, so can someone give me a hint?
 [/INST] Let $z:=\sqrt{3x-4}\ge0$. We have
$$z+\sqrt[3]{1-z^2}=1$$
or by cubing,
$$(z-1)^3+(1-z^2)=z^3-4z^2+3z=z(z-1)(z-3)=0.$$
This yields three solutions in $x$,
$$\frac43,\frac53,\frac{13}3.$$
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