1975 AHSME Problems/Problem 27: Difference between revisions
Henry wang (talk | contribs) Created page with "If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or <cmath>p^3 = p^2 - p + 2.</cmath> Similarly, <math>q^3 = q^2 - q..." |
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==Problem== | |||
If <math>p, q</math> and <math>r</math> are distinct roots of <math>x^3-x^2+x-2=0</math>, then <math>p^3+q^3+r^3</math> equals | |||
<math>\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}</math> | |||
If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or | If <math>p</math> is a root of <math>x^3 - x^2 + x - 2 = 0</math>, then <math>p^3 - p^2 + p - 2 = 0</math>, or | ||
<cmath>p^3 = p^2 - p + 2.</cmath> | <cmath>p^3 = p^2 - p + 2.</cmath> | ||
Revision as of 21:52, 12 February 2021
Problem
If 
and 
are distinct roots of 
, then 
equals
If 
is a root of 
, then 
, or
![\[p^3 = p^2 - p + 2.\]](http://latex.artofproblemsolving.com/a/b/1/ab14783187a6de9aa4e76a22ca331e47e9bea061.png)
Similarly, 
, and 
, so
![\[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]](http://latex.artofproblemsolving.com/c/2/f/c2ff0069518236fcc85c40547ec20c41a881ebdd.png)
By Vieta's formulas, 
, 
, and 
. Squaring the equation , we get
![\[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\]](http://latex.artofproblemsolving.com/6/b/3/6b3983c5f04ac79228129b8f1f75a0ee05c9d353.png)
Subtracting 
, we get
![\[p^2 + q^2 + r^2 = -1.\]](http://latex.artofproblemsolving.com/9/f/b/9fb93c5f516cc146c95ac402a1d18a2f9c3a77bf.png)
Therefore, 
. The answer is (E).
