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 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. Thanks from v8archie, Country Boy and thecoldroad 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 $x^3-x+1=0$, 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 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 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 $\displaystyle -5$. 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 Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post rakmo Algebra 3 November 22nd, 2012 11:47 PM ershi Number Theory 2 April 18th, 2012 05:29 PM MathMonster Calculus 3 November 3rd, 2011 11:30 AM ANg1.0 Calculus 0 October 30th, 2011 02:08 PM SidT Algebra 2 November 23rd, 2009 04:09 PM

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