# How would you find the exact roots of y = x 3 </msup> + x 2

How would you find the exact roots of $y={x}^{3}+{x}^{2}-2x-1$?
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Brooks Butler
Let $p\left(x\right)={x}^{3}+{x}^{2}-2x-1$, we have
$p\left(t+{t}^{-1}\right)={t}^{3}+{t}^{2}+t+1+{t}^{-1}+{t}^{-2}+{t}^{-3}=\frac{{t}^{7}-1}{{t}^{3}\left(t-1\right)}$
The RHS has roots of the form $t={e}^{±\frac{2k\pi }{7}i}$ ( coming from the ${t}^{7}-1$ factor in numerator ) for k=1,2,3. So p(x) has roots of the form
${e}^{\frac{2k\pi }{7}i}+{e}^{-\frac{2k\pi }{7}i}=2\mathrm{cos}\left(\frac{2k\pi }{7}\right)$
for k=1,2,3