# Difference between revisions of "1954 AHSME Problems/Problem 40"

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<math>\left (a+\frac{1}{a} \right )^2=3 \implies a+\frac{1}{a} =\sqrt{3}</math> | <math>\left (a+\frac{1}{a} \right )^2=3 \implies a+\frac{1}{a} =\sqrt{3}</math> | ||

− | <math>a+\frac{1}{a} = \sqrt{3} \implies (a+\frac{1}{a})^3=sqrt{ | + | <math>a+\frac{1}{a} = \sqrt{3} \implies (a+\frac{1}{a})^3=sqrt{27}</math> |

− | <math>(a+\frac{1}{a})^3=sqrt{ | + | <math>(a+\frac{1}{a})^3=sqrt{27}\implies a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27}</math> |

<math>a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27}</math> | <math>a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27}</math> |

## Revision as of 16:20, 15 April 2017

## Problem 40

If , then equals:

## Solution 1

,

## Solution 2