Find roots of polynomial $$ p(x) = x^9+3x^6+3x^3+x $$

The roots of polynomial $ p(x) $ are:

$$ \begin{aligned}x_1 &= 0\\[1 em]x_2 &= -0.0479+0.5611i\\[1 em]x_3 &= -0.0479-0.5611i\\[1 em]x_4 &= -1.2295+0.2593i\\[1 em]x_5 &= -1.2295-0.2593i\\[1 em]x_6 &= 0.4984+1.1494i\\[1 em]x_7 &= 0.4984-1.1494i\\[1 em]x_8 &= 0.7789+0.816i\\[1 em]x_9 &= 0.7789-0.816i \end{aligned} $$

Step 1:

Factor out $ \color{blue}{ x }$ from $ x^9+3x^6+3x^3+x $ and solve two separate equations:

$$ \begin{aligned} x^9+3x^6+3x^3+x & = 0\\[1 em] \color{blue}{ x }\cdot ( x^8+3x^5+3x^2+1 ) & = 0 \\[1 em] \color{blue}{ x = 0} ~~ \text{or} ~~ x^8+3x^5+3x^2+1 & = 0 \end{aligned} $$

One solution is $ \color{blue}{ x = 0 } $. Use second equation to find the remaining roots.

Step 2:

Polynomial $ x^8+3x^5+3x^2+1 $ has no rational roots that can be found using Rational Root Test, so the roots were found using Newton method.

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