# Difference between revisions of "1975 AHSME Problems/Problem 27"

(→Problem) |
|||

Line 6: | Line 6: | ||

− | + | ==Solution 1== | |

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> | ||

Line 18: | Line 18: | ||

Therefore, <math>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}</math>. The answer is (E). | Therefore, <math>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}</math>. The answer is (E). | ||

+ | ==Solution 2(Faster)== | ||

+ | We know that <math>p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr</math>. By Vieta's formulas, <math>p+q+r=1</math>,<math>pqr=2</math>, and <math>pq+qr+pr=1</math>. | ||

+ | So if we can find <math>p^2+q^2+r^2</math>, we are done. Notice that <math>(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr</math>, so <math>p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1</math>, which means that <math>p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{\text{(E)}4}</math> | ||

+ | |||

+ | ~pfalcon |

## Latest revision as of 22:03, 12 February 2021

## Problem

If and are distinct roots of , then equals

## Solution 1

If is a root of , then , or Similarly, , and , so

By Vieta's formulas, , , and . Squaring the equation , we get Subtracting , we get

Therefore, . The answer is (E).

## Solution 2(Faster)

We know that . By Vieta's formulas, ,, and . So if we can find , we are done. Notice that , so , which means that

~pfalcon