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

(→‎Solution 1)
(→‎Solution 1)
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<cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath>
 
<cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath>
  
−
By Vieta's formulas, <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get
+
By [[Vieta's formulas]], <math>p + q + r = 1</math>, <math>pq + pr + qr = 1</math>, and <math>pqr = 2</math>. Squaring the equation <math>p + q + r = 1</math>, we get
 
<cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath>
 
<cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath>
 
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get
 
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get

Revision as of 23:06, 18 March 2023

Problem

If $p, q$ and $r$ are distinct roots of $x^3-x^2+x-2=0$, then $p^3+q^3+r^3$ equals

$\textbf{(A)}\ -1 \qquad \textbf{(B)}\ 1 \qquad \textbf{(C)}\ 3 \qquad \textbf{(D)}\ 5 \qquad \textbf{(E)}\ \text{none of these}$


Solution 1

If $p$ is a root of $x^3 - x^2 + x - 2 = 0$, then $p^3 - p^2 + p - 2 = 0$, or \[p^3 = p^2 - p + 2.\] Similarly, $q^3 = q^2 - q + 2$, and $r^3 = r^2 - r + 2$, so \[p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.\]

By Vieta's formulas, $p + q + r = 1$, $pq + pr + qr = 1$, and $pqr = 2$. Squaring the equation $p + q + r = 1$, we get \[p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.\] Subtracting $2pq + 2pr + 2qr = 2$, we get \[p^2 + q^2 + r^2 = -1.\]

Therefore, $p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6 = (-1) - 1 + 6 = \boxed{4}$. The answer is $\text{(E)}$.

Solution 2(Faster)

We know that $p^3+q^3+r^3=(p+q+r)(p^2+q^2+r^2-pq-qr-pr)+3pqr$. By Vieta's formulas, $p+q+r=1$,$pqr=2$, and $pq+qr+pr=1$. So if we can find $p^2+q^2+r^2$, we are done. Notice that $(p+q+r)^2=p^2+q^2+r^2+2pq+2qr+2pr$, so $p^2+q^2+r^2=(p+q+r)^2-2(pq+qr+pr)=1^2-2\cdot1=-1$, which means that $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{\text{(E) }4}$

~pfalcon

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