Difference between revisions of "1954 AHSME Problems/Problem 40"
Katzrockso (talk | contribs) (Created page with "== Problem 40== If <math>\left (a+\frac{1}{a} \right )^2=3</math>, then <math>a^3+\frac{1}{a^3}</math> equals: <math> \textbf{(A)}\ \frac{10\sqrt{3}}{3}\qquad\textbf{(B)}\ ...") |
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<math> \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} </math> | <math> \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} </math> | ||
− | == Solution == | + | == Solution 1 == |
<math>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</math>, <math>\fbox{C}</math> | <math>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</math>, <math>\fbox{C}</math> | ||
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+ | == Solution 2 == | ||
+ | <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{3^3}</math> | ||
+ | |||
+ | <math>(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}</math> | ||
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+ | <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> | ||
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+ | <math>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}</math> | ||
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+ | <math>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}</math> | ||
+ | |||
+ | <math>3(\sqrt{3})=\sqrt{27}</math> | ||
+ | |||
+ | <math>a^3+\frac{1}{a^3}=0</math> | ||
+ | |||
+ | ==Solution 3== | ||
+ | <math>(a+\frac{1}{a})=\sqrt{3}</math> | ||
+ | |||
+ | <math>a^{2}+2+\frac{1}{a^{2}}=3 \Rightarrow a^{2}+\frac{1}{a^{2}}=1</math> | ||
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+ | <math>(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}</math> |