TITLE: Square roots and cube roots equation
QUESTION [5 upvotes]: Solve over real $a$
$$\sqrt{3a-4}+\sqrt[3]{5-3a}=1.$$
If $p=3a-4$,
$$\sqrt{p}+\sqrt[3]{1-p}=1.$$
If $q=5-3a$,
$$\sqrt{1-q}+\sqrt[3]{q}=1.$$
Seems useful, but not sure how to proceed.

REPLY [2 votes]: One strategy is to get rid of the cube root first. Two reasons: first, it is the most complicated part of the problem; second, if you are working in the real numbers you can always take a cube root, but square roots require some care.
So write $x^3=5-3a$ to get $\sqrt {1-x^3}=1-x$
(It is obvious from this that $x=1$ is a solution). Now square this, conscious that squaring is likely to produce solutions which belong to the negative square root as well as the positive square root. $$1-x^3=(1-x)^2$$
One solution is $x=1$, and otherwise divide by $(1-x)$ to obtain $1+x+x^2=1-x$ or $x^2+2x=0$
Then work backwards and check that the solutions work for the original equation (and not the one with the negative square root).
That seems to work more smoothly than the other methods which have been suggested.