# Francois Viete, France, Vietnam and Algebra: RMO training for algebra

Reference 1: https://en.wikipedia.org/wiki/Vieta%27s_formulas

Reference 2: Selected Problems of the Vietnamese Mathematical Olympiad (1962-2009), Le Hai Chau, and Le Hai Khoi, published by World Scientific;

Question:

Without solving the cubic equation, $x^{3}-x+1=0$, compute the sum of the eighth powers of all roots of the equation.

Approach: we want to be able to express the sum of the eighth powers of the three roots in terms of the three Viete’s relations here.

If $x_{1}$, $x_{2}$, $x_{3}$ are roots of the given cubic equation then, by Viete’s relations between roots and coefficients, we can say the following:

$x_{1}+x_{2}+x_{3}=0$

$x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1}=-1$

$x_{1}x_{2}x_{3}=-1$.

Furthermore, from $x_{i}^{3}-x_{i}+1=0$, it follows that

$x_{i}^{3}=x_{i}-1$

$x_{i}^{5}=x_{i}^{3}.x_{i}^{2}=(x_{i}-1)x_{i}^{2}=x_{i}^{3}-x_{i}^{2}=-x_{i}^{2}+x_{i}-1$

Then, $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=2(x_{1}^{2}+x_{2}^{2}+x_{3}^{2}) - 3(x_{1}+x_{2}+x_{3})+6$

But, $x_{1}^{2}+x_{2}^{2}+x_{3}^{2}=(x_{1}^{2}+x_{2}^{2}+x_{3}^{2})^{2}-2(x_{1}x_{2}+x_{2}x_{3}+x_{3}x_{1})=2$ and so $x_{1}^{8}+x_{2}^{8}+x_{3}^{8}=4-0+6=10$.

More later,

Nalin Pithwa.

This site uses Akismet to reduce spam. Learn how your comment data is processed.