
PreCalculus PreCalculus Math Forum 
 LinkBack  Thread Tools  Display Modes 
June 21st, 2017, 02:42 AM  #1 
Newbie Joined: Jun 2017 From: Earth Posts: 3 Thanks: 0  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! 
June 21st, 2017, 06:27 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
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. 
June 21st, 2017, 07:51 AM  #3 
Math Team Joined: Dec 2013 From: Colombia Posts: 7,674 Thanks: 2654 Math Focus: Mainly analysis and algebra 
If $p$ is a root of $x^3x+1=0$, then $p^3p+1=0 \implies p^3=p1$. Similarly for $q$ and $r$. Then, $p^5=\frac{p^6}p=\frac{(p1)^2}p = \frac{p^22p+1}p=p2+\frac1p$. 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. 
June 21st, 2017, 10:22 AM  #4 
Newbie Joined: Jun 2017 From: Earth Posts: 3 Thanks: 0 
Thanks guys, helps a lot!

June 21st, 2017, 11:31 AM  #5 
Senior Member Joined: Feb 2010 Posts: 711 Thanks: 147 
I know this is a precalculus 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 $\displaystyle f(x)=x^3x+1$ this simplifies to $\displaystyle \dfrac{3x^2}{1x^2+x^3}$. Now do a long division to obtain the infinite series $\displaystyle 3+0x+2x^23x^3+2x^45x^5+ \cdots$ In this series, the coefficient of $\displaystyle x^n$ is $\displaystyle p^n+q^n+r^n$, so as said earlier, the answer is $\displaystyle 5$. 
June 21st, 2017, 04:51 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,919 Thanks: 2201 
As $x^5 + x^2  x + 1 = (x^3  x + 1)(x^2 + 1) = 0$, $\begin{align*}p^5 + q^5 + r^5 &= (p^2 + q^2 + r^2) + (p + q + r)  3 \\ &= (p + q + r)^2 + 2(pq + pr + qr) + p + q + r  3 \\ &= 2  3 \\ &= 5.\end{align*}$ 

Tags 
cubic, roots, solutions, zeros 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
cubic  rakmo  Algebra  3  November 22nd, 2012 11:47 PM 
mod of a cubic  ershi  Number Theory  2  April 18th, 2012 05:29 PM 
Help in this cubic equation..  MathMonster  Calculus  3  November 3rd, 2011 11:30 AM 
Cubic Spline  ANg1.0  Calculus  0  October 30th, 2011 02:08 PM 
Solutions to a Cubic  SidT  Algebra  2  November 23rd, 2009 04:09 PM 