1975 AHSME Problems/Problem 27: Difference between revisions

Latest revision as of 17:58, 23 August 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

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

@@ Line 12: / Line 12: @@
 <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>
 Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get
@@ Line 24: / Line 24: @@
 ~pfalcon
+==Solution 3 (Beginner's Solution)==
+Use Vieta's formulas to get <math>p+q+r=1</math>, <math>pq+qr+pr=1</math>, and <math>pqr=2</math>.
+Square <math>p+q+r=1</math>, and get <math>p^2+q^2+r^2+2pq+2pr+2qr=1</math>
+Substitute <math>pq+qr+pr=1</math> and simplify to get <math>-1=p^2+q^2+r^2</math>
+After that, multiply both sides by <math>1=p+q+r</math>, to get
+<math>-1=p^3+q^3+r^3+p^2q+q^2r+p^2r+q^2r+r^2p+r^2q</math>
+Then, factor out <math>pq</math>, <math>qr</math>, and <math>pr</math>:
+<math>-1=p^3+q^3+r^3+pq(p+q)+qr(q+r)+pr(p+r)</math>
+Then, substitute the first equation into <math>p+q</math>, <math>q+r</math>, and <math>p+r</math>.
+<math>-1=p^3+q^3+r^3+pq(1-r)+qr(1-p)+pr(1-q)</math>
+Then, multiply it out:
+<math>-1=p^3+q^3+r^3+pq+qr+pr-3pqr</math>
+After that, substitute the equations <math>pq+qr+pr=1</math> and <math>pqr=2</math>:
+<math>-1=p^3+q^3+r^3+1-6</math>
+Solving that, you get <math>p^3+q^3+r^3=\boxed{\text{(E) }4}</math>
+~EZ PZ Ms.Lemon SQUEEZY

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