The solution set of the equation (2y−3)13−(4y+5)13=0

Step 1
Consider the provided equation,
${\displaystyle {(2y-3)}^{\frac{1}{3}}-{(4y+5)}^{\frac{1}{3}}=0}$
Now, add ${\displaystyle {\left(4y_5\right)}^{\frac{1}{3}}}$ to the both sides,
${\displaystyle {(2y-3)}^{\frac{1}{3}}-{(4y+5)}^{\frac{1}{3}}=0}$
${\displaystyle {(2t-3)}^{\frac{1}{3}}-{(4y+5)}^{\frac{1}{3}}+{(4y+5)}^{\frac{1}{3}}={(4y+5)}^{\frac{1}{3}}}$
${\displaystyle {(2y-3)}^{\frac{1}{3}}={(4y+5)}^{\frac{1}{3}}}$
Take cube on both sides of the above equation,
${\displaystyle {(2y-3)}^{\frac{1}{3}}={(4y+5)}^{\frac{1}{3}}}$
${\displaystyle {\left({(2y-3)}^{\frac{1}{3}}\right)}^3={\left({(4y+5)}^{\frac{1}{3}}\right)}^3}$
${\displaystyle 2y-3=4y+5}$
${\displaystyle -3-5=4y-2y}$
Simplify the above equation,
${\displaystyle -3-5=4y-2y}$
${\displaystyle -8=2y}$
${\displaystyle y=-4}$
Check:
Now, substitute the value of ${\displaystyle y=-4}$ in the provided equation $(2y-3{)}^{\frac{1}{3}}-(4y+5{)}^{\frac{1}{3}}=0$
$[2(-4)-3{]}^{\frac{1}{3}}-[4(-4)+5{]}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$(-8-3{)}^{\frac{1}{3}}-(-16+5{)}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$(-11{)}^{\frac{1}{3}}-(-11{)}^{\frac{1}{3}}\begin{array}{c}?\\ =\end{array}0$
$0\begin{array}{c}?\\ =\end{array}0$
Which is true.
Hence, the solution set of the equation ${\displaystyle {(2y-3)}^{\frac{1}{3}}-{(4y+5)}^{\frac{1}{3}}=0}$ is ${\displaystyle \{-4\}}$