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1954 AHSME Problems/Problem 40

Revision as of 23:21, 26 April 2020 by Fortytwok (talk | contribs) (→‎Solution 3)
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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

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

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

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

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

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

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

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

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

Solution 3

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

$a^{2}+2+\frac{1}{a^{2}}=3 \Rightarrow a^{2}+\frac{1}{a^{2}}=1$

$(a^{3}+\frac{1}{a^{3}})=(a+\frac{1}{a})(a^{2}-1+\frac{1}{a^{2}}) \Rightarrow (a^{3}+\frac{1}{a^{3}})=\sqrt{3}(1-1) \Rightarrow \fbox{C}$

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