$$4y(4y^2-7y+1) = 0$$
Answer
$$ \begin{matrix}y_1 = 0 & y_2 = \dfrac{ 7 }{ 8 }-\dfrac{\sqrt{ 33 }}{ 8 } & y_3 = \dfrac{ 7 }{ 8 }+\dfrac{\sqrt{ 33 }}{ 8 } \end{matrix} $$
Explanation
$$ \begin{aligned} 4y(4y^2-7y+1) &= 0&& \text{simplify left side} \\[1 em]16y^3-28y^2+4y &= 0&& \\[1 em] \end{aligned} $$
In order to solve $ \color{blue}{ 16x^{3}-28x^{2}+4x = 0 } $, first we need to factor our $ x $.
$$ 16x^{3}-28x^{2}+4x = x \left( 16x^{2}-28x+4 \right) $$$ x = 0 $ is a root of multiplicity $ 1 $.
The remaining roots can be found by solving equation $ 16x^{2}-28x+4 = 0$.
$ 16x^{2}-28x+4 = 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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