Find the sum of cubes of roots of a biquadratic

Question

Given that $a,b,c,d$ are the roots of the equation $x^4-3x^3+x^2-2x+1=0$, find the value of $a^3+b^3+c^3+d^3$. Since, there are no 'zero' coefficients in the equation, it looks like a tough job for Newton's Sums. Is there some trick? Thanks

Answer 1

As $abcd\ne0,$

$$a^4-3a^3+a^2-2a+1=0\iff a^3-3a^2+a-2+\dfrac1a=0$$

$$\sum a^3=3\sum a^2-\sum a+2\sum1-\sum\dfrac1a$$

Now by Vieta's formula, $\sum a=3$ and $\sum ab=1,\sum abc=2,abcd=1$

$\implies\sum a^2=(\sum a)^2-2\sum ab=\cdots$

Finally, $\sum\dfrac1a=\dfrac{\sum abc}{abcd}=\cdots$

Answer 2

Hint: rather than a trick, there's a theorem, due to Newton: it says that a symmetric (polynomial) function $q(x_1, \ldots, x_n)$ of the roots of a polynomial $p$ of degree $n$ is a polynomial function of the coefficients of $p$. See the wikipedia pages on symmetric polynomials and Newton's identities for details.