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

(→‎Solution 2)
(→‎Solution 2)
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<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 02:53, 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
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All AHSME Problems and Solutions

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