1954 AHSME Problems/Problem 40

Revision as of 16:20, 15 April 2017 by Xiej (talk | contribs) (Solution 2)

Problem 40

If $\left (a+\frac{1}{a} \right )^2=3$, then $a^3+\frac{1}{a^3}$ equals:

$\textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ 3\sqrt{3}\qquad\textbf{(C)}\ 0\qquad\textbf{(D)}\ 7\sqrt{7}\qquad\textbf{(E)}\ 6\sqrt{3}$

Solution 1

$a+\frac{1}{a}=\sqrt{3}\implies (a+\frac{1}{a})^3=3\sqrt{3}\implies a^3+3\frac{a^2}{a}+\frac{3a}{a^2}+\frac{1}{a^3}\implies a^3+3(a+\frac{1}{a})+\frac{1}{a^3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}+3\sqrt{3}=3\sqrt{3}\implies a^3+\frac{1}{a^3}=0$, $\fbox{C}$

Solution 2

$\left (a+\frac{1}{a} \right )^2=3 \implies a+\frac{1}{a} =\sqrt{3}$

$a+\frac{1}{a} = \sqrt{3} \implies (a+\frac{1}{a})^3=sqrt{27}$

$(a+\frac{1}{a})^3=sqrt{27}\implies a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27}$

$a^3+3a^2\frac{1}{a}+3a\frac{1}{a^2}+\frac{1}{a^3}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27}$

$a^3+\frac{1}{a^3}+3a+\frac{3}{a}=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27}$

$a^3+\frac{1}{a^3}+3(a+\frac{1}{a})=\sqrt{27} \implies a^3+\frac{1}{a^3}+3(\sqrt{3})=\sqrt{27}$

$3(\sqrt{3})=\sqrt{27}$

$a^3+\frac{1}{a^3}=0$

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