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

~pfalcon

## Solution 3 (Beginner's Solution)

Use Vieta's formulas to get $p+q+r=1$, $pq+qr+pr=1$, and $pqr=2$.

Square $p+q+r=1$, and get $p^2+q^2+r^2+2pq+2pr+2qr=1$

Substitute $pq+qr+pr=1$ and simplify to get $-1=p^2+q^2+r^2$

After that, multiply both sides by $1=p+q+r$, to get $-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q$

Then, factor out $pq$, $qr$, and $pr$: $-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)$

Then, substitute the first equation into $p+q$, $q+r$, and $p+r$. $-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)$

Then, multiply it out: $-1=p^3+q^3+r^3+pq+qr+pr-3pqr$

After that, substitute the equations $pq+qr+pr=1$ and $pqr=2$: $-1=p^3+q^3+r^3+1-6$

Solving that, you get $p^3+q^3+r^3=\boxed{\text{(E) }4}$ ~EZ PZ Ms.Lemon SQUEEZY