My Math Forum  

Go Back   My Math Forum > High School Math Forum > Pre-Calculus

Pre-Calculus Pre-Calculus Math Forum


Thanks Tree6Thanks
  • 3 Post By skipjack
  • 3 Post By v8archie
Reply
 
LinkBack Thread Tools Display Modes
June 21st, 2017, 03:42 AM   #1
Newbie
 
Joined: Jun 2017
From: Earth

Posts: 3
Thanks: 0

Question Sum of ^5 of cubic solutions

I came across this problem, and I'm only confident we shouldn't try to actually solve for roots of x. The solution is 5, but I don't know how to approach the problem. I'm guessing it has something to do with matrices.

Problem:
"If p, q and r are roots of x^3 - x + 1=0, then p^5 + q^5 + r^5=?"

Any help is greatly appreciated, thanks!
thecoldroad is offline  
 
June 21st, 2017, 07:27 AM   #2
Global Moderator
 
Joined: Dec 2006

Posts: 18,164
Thanks: 1423

Let the zeros of x³ - x + 1 be p, q and r.
Let a = p + q + r, b = pq + pr + qr, and c = pqr.
By substituting these expressions for a, b and c, a^5 - 5(ab - c)(a² - b) ≡ p$^5$ + q$^5$ + r$^5\!$.

As x³ - x + 1 ≡ (x - p)(x - q)(x - r) ≡ x³ - ax² + bx - c,
equating coefficients gives a = 0, b = -1 and c = -1.

Hence p$^5$ + q$^5$ + r$^5\!$ = 0$^5$ - 5(0 + 1)(1) = -5.
skipjack is offline  
June 21st, 2017, 08:51 AM   #3
Math Team
 
Joined: Dec 2013
From: Colombia

Posts: 7,042
Thanks: 2344

Math Focus: Mainly analysis and algebra
If $p$ is a root of $x^3-x+1=0$, then $p^3-p+1=0 \implies p^3=p-1$. Similarly for $q$ and $r$.

Then, $p^5=\frac{p^6}p=\frac{(p-1)^2}p = \frac{p^2-2p+1}p=p-2+\frac1p$. And similarly for $q$ and $r$.

Now $p^5 + q^5 + r^5=(p+q+r) - 6 + (\frac1p + \frac1q + \frac1r)= (p+q+r) - 6 + \frac{qr+rp+pq}{pqr}$

As Skipjack pointed out, the first term, and the numerator and denominator of the last term are three coefficients of the original equation and thus we can recover their numerical values.
Thanks from EvanJ, Country Boy and thecoldroad
v8archie is offline  
June 21st, 2017, 11:22 AM   #4
Newbie
 
Joined: Jun 2017
From: Earth

Posts: 3
Thanks: 0

Thanks guys, helps a lot!
thecoldroad is offline  
June 21st, 2017, 12:31 PM   #5
Senior Member
 
mrtwhs's Avatar
 
Joined: Feb 2010

Posts: 632
Thanks: 103

I know this is a pre-calculus forum but if you know the very basics of finding a derivative, there is another way to do this.

Form the fraction $\displaystyle \dfrac{f^{\prime}(1/x)}{x \cdot f(1/x)}$.

With $\displaystyle f(x)=x^3-x+1$ this simplifies to $\displaystyle \dfrac{3-x^2}{1-x^2+x^3}$.

Now do a long division to obtain the infinite series

$\displaystyle 3+0x+2x^2-3x^3+2x^4-5x^5+ \cdots$

In this series, the coefficient of $\displaystyle x^n$ is $\displaystyle p^n+q^n+r^n$, so as said earlier, the answer is $\displaystyle -5$.
mrtwhs is offline  
June 21st, 2017, 05:51 PM   #6
Global Moderator
 
Joined: Dec 2006

Posts: 18,164
Thanks: 1423

As $x^5 + x^2 - x + 1 = (x^3 - x + 1)(x^2 + 1) = 0$,
$\begin{align*}p^5 + q^5 + r^5 &= -(p^2 + q^2 + r^2) + (p + q + r) - 3 \\
&= -(p + q + r)^2 + 2(pq + pr + qr) + p + q + r - 3 \\
&= -2 - 3 \\
&= -5.\end{align*}$
skipjack is offline  
Reply

  My Math Forum > High School Math Forum > Pre-Calculus

Tags
cubic, roots, solutions, zeros



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
cubic rakmo Algebra 3 November 23rd, 2012 12:47 AM
mod of a cubic ershi Number Theory 2 April 18th, 2012 06:29 PM
Help in this cubic equation.. MathMonster Calculus 3 November 3rd, 2011 12:30 PM
Cubic Spline ANg1.0 Calculus 0 October 30th, 2011 03:08 PM
Solutions to a Cubic SidT Algebra 2 November 23rd, 2009 05:09 PM





Copyright © 2017 My Math Forum. All rights reserved.