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Difference between revisions of "1975 AHSME Problems/Problem 27"
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==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>
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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).
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==Solution 2(Faster)==
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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>.
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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>           
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~pfalcon

Revision as of 22:03, 12 February 2021

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 (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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