Technique to simplify algebraic calculations on roots of polynomial

Question

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.

lab bhattacharjee · Answer 1 · 2014-05-11T05:47:31.053

1

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

edited May 11 '14 at 05:47

answered May 11 '14 at 05:40

lab bhattacharjee

274,582

I know that...But it would get harder to do this when power is more...I want to instead know the technique of forming a new quadratic related to what we want to find – Shaurya Gupta May 11 '14 at 05:44
@shauryagupta, Find the edited version – lab bhattacharjee May 11 '14 at 05:48
It's still not what the technique was. you don't have to rearrange the given equation. It really simplified things a lot...You need to make a completely new quadratic. – Shaurya Gupta May 11 '14 at 05:55
What if I were to find $\alpha^3+\beta^3+\gamma^3$ in $t^3-9t^2+26t-24$? – Shaurya Gupta May 23 '14 at 04:41
@shauryagupta, I think this comment was meant for http://math.stackexchange.com/questions/806201/number-of-solutions-to-the-equations-x-2y-4z-9-4yz-2xz-xy-13-xyz and the solution pattern can be http://math.stackexchange.com/questions/806261/prove-3-3-s3-3ps-in-quadratic-equation/806290#806290 – lab bhattacharjee May 23 '14 at 06:25
No, this comment is meant to be here. And the technique I'm looking for is very useful for this. what would be an equation whose roots are $\alpha^3, \beta^3 \text{ and } \gamma^3$? – Shaurya Gupta May 24 '14 at 04:01
@shauryagupta, But $\gamma$ is not defined in the Question, itself? – lab bhattacharjee May 24 '14 at 04:17
It is the third root of the cubic polynomial $t^3 - 9t^2 + 26t - 24$. – Shaurya Gupta May 24 '14 at 04:44
@shauryagupta, Where is the cubic Polynomial ? – lab bhattacharjee May 24 '14 at 04:51
It's $t^3−9t^2+26t−24$. – Shaurya Gupta May 24 '14 at 04:52
Let us continue this discussion in chat. – lab bhattacharjee May 24 '14 at 04:54

WimC · Answer 2 · 2014-05-11T06:47:41.923

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 $X^2-123 X+1$.

score 0 · Answer 3 · answered May 11 '14 at 06:34

0

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.

answered May 11 '14 at 06:34

Caleb Stanford

45,895

please see my updated question. – Shaurya Gupta May 11 '14 at 08:41
1

7 years later I found this answer extremely useful. – Nov 01 '21 at 18:46
@arthur Good to hear! :) – Caleb Stanford Nov 01 '21 at 19:15

score 0 · Answer 4 · answered May 11 '14 at 14:34

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.

Technique to simplify algebraic calculations on roots of polynomial

4 Answers4