Find roots of polynomial $$ p(x) = y^6-2y^3 $$
Answer
The roots of polynomial $ p(y) $ are:
$$ \begin{aligned}y_1 &= 0\\[1 em]y_2 &= 1.2599\\[1 em]y_3 &= -0.63+1.0911i\\[1 em]y_4 &= -0.63-1.0911i \end{aligned} $$Explanation
Step 1:
Factor out $ \color{blue}{ y^3 }$ from $ y^6-2y^3 $ and solve two separate equations:
$$ \begin{aligned} y^6-2y^3 & = 0\\[1 em] \color{blue}{ y^3 }\cdot ( y^3-2 ) & = 0 \\[1 em] \color{blue}{ y^3 = 0} ~~ \text{or} ~~ y^3-2 & = 0 \end{aligned} $$One solution is $ \color{blue}{ y = 0 } $. Use second equation to find the remaining roots.
Step 2:
Polynomial $ y^3-2 $ has no rational roots that can be found using Rational Root Test, so the roots were found using qubic formulas.
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