$$16y^3-25y = 0$$
Answer
$$ \begin{matrix}y_1 = 0 & y_2 = \dfrac{ 5 }{ 4 } & y_3 = -\dfrac{ 5 }{ 4 } \end{matrix} $$
Explanation
In order to solve $ \color{blue}{ 16x^{3}-25x = 0 } $, first we need to factor our $ x $.
$$ 16x^{3}-25x = x \left( 16x^{2}-25 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 16x^{2}-25 = 0$.
$ 16x^{2}-25 = 0 $ is a quadratic equation.
You can use step-by-step quadratic equation solver to see a detailed explanation on how to solve this equation.
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