1963 AHSME Problems/Problem 40: Difference between revisions

Latest revision as of 13:03, 26 March 2023

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,  . by solving this, we get 
. this gives the step what we had done in solution 1.The answer is .

Solution 3

Cubing both sides, we get $x+9-3\sqrt[3]{(x+9)^2(x-9)}+3\sqrt[3]{(x+9)(x-9)^2}-x+9=27,$ so $18-3\sqrt[3]{(x-9)(x+9)}(\sqrt[3]{x+9}-\sqrt[3]{x-9})=27\implies 3\sqrt[3]{x^2-81}=-3.$ Dividing both sides by 3 and cubing, we find $x^2=80$ , which is between $\boxed{(C)75\text{ and }85}$ .

~pfalcon

Solution 4

We consider the formula $(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ . Factoring out $3ab$ , and the commutative property gives $( a^3 - b^3) - 3ab(a-b)$ . Cubing both sides gives us $(\sqrt[3]{x+9}-\sqrt[3]{x-9})^3=27$ Using our formula, we have $18 - 9 \sqrt[3]{x^2 - 81} = 27$ Solving this gives $x^2 = 80$ , therefore, the answer is $\boxed{\textbf{(C)}}$ . ~zixuan12

@@ Line 21: / Line 21: @@
 ==Solution 2==
 <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>.
+==Solution 3==
+Cubing both sides, we get
+<cmath>x+9-3\sqrt[3]{(x+9)^2(x-9)}+3\sqrt[3]{(x+9)(x-9)^2}-x+9=27,</cmath>so <cmath>18-3\sqrt[3]{(x-9)(x+9)}(\sqrt[3]{x+9}-\sqrt[3]{x-9})=27\implies 3\sqrt[3]{x^2-81}=-3.</cmath> Dividing both sides by 3 and cubing, we find <math>x^2=80</math>, which is between <math>\boxed{(C)75\text{ and }85}</math>.
+~pfalcon
+==Solution 4==
+We consider the formula <math>(a-b)^3 = a^3 - 3a^2b + 3ab^2 - b^3</math>. Factoring out <math>3ab</math>, and the commutative property gives <math>(
+a^3 - b^3) - 3ab(a-b)</math>. Cubing both sides gives us
+<cmath>(\sqrt[3]{x+9}-\sqrt[3]{x-9})^3=27</cmath>
+Using our formula, we have
+<cmath>18 - 9 \sqrt[3]{x^2 - 81} = 27</cmath>
+Solving this gives <math>x^2 = 80</math>, therefore, the answer is <math>\boxed{\textbf{(C)}}</math>.
+~zixuan12
 ==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

Page

Toolbox

Search