Art of Problem Solving
Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

1975 AHSME Problems/Problem 27: Difference between revisions

← Older edit
Pfalcon (talk | contribs)
editor
12 edits
No edit summary
fix mistake
 
(6 intermediate revisions by 3 users not shown)
Line 6: Line 6:




 
==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>
Line 12: Line 12:
<cmath>p^3 + q^3 + r^3 = (p^2 + q^2 + r^2) - (p + q + r) + 6.</cmath>
<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>
<cmath>p^2 + q^2 + r^2 + 2pq + 2pr + 2qr = 1.</cmath>
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get
Subtracting <math>2pq + 2pr + 2qr = 2</math>, we get
<cmath>p^2 + q^2 + r^2 = -1.</cmath>
<cmath>p^2 + q^2 + r^2 = -1.</cmath>


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 <math>\text{(E)}</math>.
 
==Solution 2(Faster)==
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>.
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>           
 
~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

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

Retrieved from "https://wiki.aopstest.com/wiki/index.php?title=1975_AHSME_Problems/Problem_27&oldid=197354"