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;

Amazon India link:

https://www.amazon.in/Selected-Problems-Vietnamese-Olympiad-Mathematical/dp/9814289590/ref=sr_1_1?s=books&ie=UTF8&qid=1510149044&sr=1-1&keywords=selected+problems+of+the+vietnamese+mathematical+olympiad

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. 

Answer:

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.

 

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