Help solve this equation: $\sqrt{3x-4}+\sqrt[3]{5-3x}=1$

Question

$\sqrt{3x-4}+\sqrt[3]{5-3x}=1$ Find x.

I've been assigned this question for homework in an online class and I'm a little bit stuck. I tried defining a variable $a=3x-9/2$ but it didn't really bring me anywhere. Hints would be greatly appreciated!

Answer 1

Hint:

We have $a+b=1$ where $a=\sqrt{3x-4}\ge0$

$a^2+b^3=3x-4+5-3x=1$

$\implies(1-b)^2+b^3=1$

Answer 2

Similar to lab bhattacharjee's answer, let $y=3x-4$ to make $$\sqrt y+ \sqrt[3]{1-y}=1$$ Since $y$ must be positive, let $y=z^2$ to make $$z+\sqrt[3]{1-z^2}=1\implies \sqrt[3]{1-z^2}=1-z\implies 1-z^2=(1-z)^3$$ that is to say $$(1-z)(1+z)=(1-z)^3$$ If $z \neq 1$, then $$1+z=(1-z)^2\implies z(z-3)=0$$ So, the roots for $z$ are $0,1,3$; then the roots for $y$ and then the roots for $x$.