My first ever problem

Algebra Level 3

Consider a polynomial $P(x)$ such that $P(x)=\displaystyle \sum _{ n=1 }^{ 7 }{ \phi (n)x^{ n } }$ and $x_{ 1 },x_{ 2 },x_{ 3 }, ...,x_{ 7 }$ are the roots of $P(x)=0$ .

Find $\displaystyle \sum _{ n=1 }^{ 7 }{ x_{ n }^{ 3 }}$ .

The answer is in the form $-\frac { p }{ q }$ , where $p$ and $q$ are co-prime positive integers, find $p+q$ .

Note : $\phi (n)$ denotes Euler's totient function.

The answer is 37.

1 solution

Otto Bretscher
Jan 4, 2016

Nice problem! We are looking forward to many more.

We have $P(x)=6x^7+2x^6+4x^5+2x^4+...$ , with the elementary symmetric functions $e_1=-\frac{2}{6},$ $e_2=\frac{4}{6},e_3=-\frac{2}{6}$ . Now we have the Girard sum $p_3=e_1^3-3e_1e_2+3e_3=-\frac{10}{27}$ so the answer is $\boxed{37}$ .

same way(+1)... is it girard sum or newton sum or is girard another name for this?

Aareyan Manzoor - 5 years, 5 months ago

The formula for the sum of the cubes was published by the French mathematician Albert Girard long before Newton was born.

Otto Bretscher - 5 years, 5 months ago

My first ever problem

The answer is 37.

1 solution

0 pending reports