Technique to simplify algebraic calculations on roots of polynomial I was once told about a technique to simplify algebra on the roots of a polynomial.
So if you want to find $\alpha^3+\beta^3+\gamma^3$, where $\alpha,\beta \text{ and } \gamma$ are roots of $ax^3+bx^2+cx+d$, then you find a new quadratic with its roots as $\alpha^3, \beta^3 \text{ and } \gamma^3$ or something like that(I don't remember exactly). Does somebody know what this technique exactly is?

Note: Please modify the question title and details as necessary.
 A: HINT:
$$\alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta)$$
$$\alpha^3\beta^3=(\alpha\beta)^3$$

Alternatively,
$$ax^2+bx+c=0\implies ax^2+bx=-c\implies (ax^2+bx)^3=(-c)^3$$
$$a^3(x^3)^2+b^3(x^3)+3ab(x^3)(-c)=-c^3$$
Replace $x^3$ with $y$ and rearrange
A: The following computes powers of roots of quadratic equations very efficiently.  It can easily be extended to equations of higher degree.  Let $\alpha$ be a root of $$X^2+a X + b.$$
In the basis $\langle 1, \alpha\rangle$ multiplication by $\alpha$ in the field $\mathbb{Q}(\alpha)$ has matrix
$$A=\begin{pmatrix}0&-b\\1&-a\end{pmatrix}.$$
Powers of $\alpha$ can be expressed in this basis by taking the same power of $A$. The coefficients of $1$ and $\alpha$ are then given by the first column of the result. Powers of $A$ can be computed by the repeated squaring method. For example, consider the quadratic $X^2-X-1$. Then $$A=\begin{pmatrix}0&1\\1&1\end{pmatrix}.$$ To compute $\alpha^{10}$ first compute $$A^{10}=\begin{pmatrix}34&55\\55&89\end{pmatrix}.$$
Now the first column shows that $\alpha^{10}=55\alpha+34$. If $\beta$ is the other root then the same argument shows that $\beta^{10}=55\beta+34$. Then
$$\begin{eqnarray}\alpha^{10}+\beta^{10}&=&55(\alpha+\beta)+68=123\\
\alpha^{10}\beta^{10}&=&(\alpha \beta)^{10}=1
\end{eqnarray}$$
and $\alpha^{10}$, $\beta^{10}$ are roots of $X^2-123 X+1$.
A: It sounds to me like you are looking for Newton's identities.
These are the general form of lab's answer.  In particular you can express power sums directly in terms of the coefficients of the polynomial. 
A: New question $\rightarrow$ new answer.  If $\alpha$, $\beta$ are the roots of $X^2+a X+ b$ then a polynomial expression $P(\alpha, \beta)$ can be reduced to a linear expression in $\alpha$ and $\beta$: First eliminate powers of $\alpha \beta = b$ from each monomial. Then use the technique from my other answer to reduce the remaining powers of $\alpha$ and $\beta$. This results in $$P(\alpha, \beta) = c_0 + c_1\alpha + c_2\beta$$ for some coefficients $c_0, c_1, c_2$. The quadratic for $P(\alpha, \beta)$ becomes $$(X-c_0)^2+a(c_1+c_2)(X-c_0) +a^2 c_1 c_2+b\, (c_1-c_2)^2.$$
Alternatively use Gröbner basis reduction on the ideal $$\langle X+Y+a, XY-b, P(X,Y)-Z \rangle$$ with $Z < X \wedge Y$ in the monomial order to find a quadratic equation in $Z$ alone. As an example consider $X^2-X-1$ and $P(\alpha, \beta) = 2\alpha + \beta$. Since this is already linear the formula above results in the following quadratic for $P(\alpha, \beta)$:
$$X^2-3X+1.$$
The same result using Gröbner basis reduction can be found here.
