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

Difference between revisions of "1963 AHSME Problems/Problem 40"

(→‎Solution 2)
(→‎Solution 2)
Line 22: Line 22:
 
<math>\sqrt[3]{x+9}-\sqrt[3]{x-9}-3=0</math> i.e, <math>\sqrt[3]{x+9}+\sqrt[3]{-x+9}+(-3)=0</math>
 
<math>\sqrt[3]{x+9}-\sqrt[3]{x-9}-3=0</math> i.e, <math>\sqrt[3]{x+9}+\sqrt[3]{-x+9}+(-3)=0</math>
 
   if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.
 
   if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.
−
  then, <math>x+9-x+9+27=3(\sqrt[3]{x+9})(\sqrt[3]{9-x})(3)</math> . by solving this, we get <math>-9=-9\sqrt[3]{9^2-x^2}</math>
+
  then, <math>x+9-x+9+27=3(\sqrt[3]{x+9})(\sqrt[3]{9-x})(-3)</math> . by solving this, we get <math>-9=-9\sqrt[3]{9^2-x^2}</math>
 
  <math>x^2=80</math>. this gives the step what we had done in solution 1.The answer is <math>\boxed{\textbf{(C)}}</math>.
 
  <math>x^2=80</math>. this gives the step what we had done in solution 1.The answer is <math>\boxed{\textbf{(C)}}</math>.
  

Revision as of 03:53, 19 March 2020

Problem

If $x$ is a number satisfying the equation $\sqrt[3]{x+9}-\sqrt[3]{x-9}=3$, then $x^2$ is between:

$\textbf{(A)}\ 55\text{ and }65\qquad \textbf{(B)}\ 65\text{ and }75\qquad \textbf{(C)}\ 75\text{ and }85\qquad \textbf{(D)}\ 85\text{ and }95\qquad \textbf{(E)}\ 95\text{ and }105$

Solution 1

Let $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$. Cubing these equations, we get $a^3 = x + 9$ and $b^3 = x - 9$, so $a^3 - b^3 = 18$. The left-hand side factors as \[(a - b)(a^2 + ab + b^2) = 18.\]

However, from the given equation $\sqrt[3]{x + 9} - \sqrt[3]{x - 9} = 3$, we get $a - b = 3$. Then $3(a^2 + ab + b^2) = 18$, so $a^2 + ab + b^2 = 18/3 = 6$.

Squaring the equation $a - b = 3$, we get $a^2 - 2ab + b^2 = 9$. Subtracting this equation from the equation $a^2 + ab + b^2 = 6$, we get $3ab = -3$, so $ab = -1$. But $a = \sqrt[3]{x + 9}$ and $b = \sqrt[3]{x - 9}$, so $ab = \sqrt[3]{(x + 9)(x - 9)} = \sqrt[3]{x^2 - 81}$, so $\sqrt[3]{x^2 - 81} = -1$. Cubing both sides, we get $x^2 - 81 = -1$, so $x^2 = 80$. The answer is $\boxed{\textbf{(C)}}$.


Solution 2

$\sqrt[3]{x+9}-\sqrt[3]{x-9}-3=0$ i.e, $\sqrt[3]{x+9}+\sqrt[3]{-x+9}+(-3)=0$ 

 if the sum of three numbers is zero, then their sum of cubes is thrice the product of each number.
then, $x+9-x+9+27=3(\sqrt[3]{x+9})(\sqrt[3]{9-x})(-3)$ . by solving this, we get $-9=-9\sqrt[3]{9^2-x^2}$
$x^2=80$. this gives the step what we had done in solution 1.The answer is $\boxed{\textbf{(C)}}$.

See Also

1963 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 39 		Followed by
Last Problem
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1963_AHSME_Problems/Problem_40&oldid=119772"