# 1975 AHSME Problems/Problem 27

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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).