<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] By $x=\frac {y^3} 3+\frac 53$
$$\sqrt{3x - 4} + \sqrt[3]{5 - 3x} = 1 \iff \sqrt {y^3+1}-y=1 $$
$$\iff y^3+1=(1+y)^2 \iff y^3-y^2-2y=0 \iff y\in \{-1,0,2\}$$
that is $x\in \{\frac43,\frac53,\frac{13}3\}$.
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