Difference between revisions of "1954 AHSME Problems/Problem 40"
(→Solution 2) |
(→Solution 3) |
||
(2 intermediate revisions by 2 users not shown) | |||
Line 11: | Line 11: | ||
<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{3^3}</math> |
− | <math>(a+\frac{1}{a})^3=sqrt{ | + | <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> |
<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> | ||
Line 24: | Line 24: | ||
<math>a^3+\frac{1}{a^3}=0</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> | ||
+ | |||
+ | <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> |