1975 AHSME Problems/Problem 27

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 $p^3+q^3+r^3=1\cdot-2+3\cdot2=\boxed{\text{(E) }4}$