Question: Let $P = 27+108x+90x^2-80x^3-60x^4+48x^5-8x^6$, find $\sqrt[3]{P}$.
Just wondering if there is a general way for dealing with this sort of question, I was able to figure how to find $\sqrt{P}$, but $\sqrt[3]{P}$ is beyond me. Long division is used in an example, unfortunately it lacks justification.
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$\begingroup$The hope is that $\sqrt[3]{P}$ is also a polynomial. If $\sqrt[3]{P}$ were to be a polynomial, then $\sqrt[3]{P}$ should be of the form $$\sqrt[3]{P} = (-2x^2 + ax + 3)$$ since the coefficient of $x^6$ is $-8$ and the constant term is $27$. Can you take this further?
$\endgroup$ 3 $\begingroup$Let $(3+ax-2x^2)^3=27+108x+90x^2-80x^3-60x^4+48x^5-8x^6$
Now use this and compare the coefficients of the different powers of $x$ to find the unique value of $a$ such that all the equation formed by comparing the coefficients remain consistent.
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