Question : If $k^4+\frac{1}{k^4}=47$, then what is the value of $k^3+\frac{1}{k^3}$?
Option 1: 4.5
Option 2: 54
Option 3: 18
Option 4: 9
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Correct Answer: 18
Solution : Given, $k^4+\frac{1}{k^4}=47$ Adding 2 on both sides, ⇒ $k^4+\frac{1}{k^4}+2=47+2$ ⇒ $(k^2)^2+\frac{1}{(k^2)^2}+2\times k^2\times \frac{1}{k^2}=49$ ⇒ $(k^2+\frac{1}{k^2})^2=49$ ⇒ $k^2+\frac{1}{k^2}=7$ Again adding 2 on both sides, $k^2+\frac{1}{k^2}+2=7+2$ ⇒ $(k+\frac{1}{k})^2=9$ ⇒ $k+\frac{1}{k}=3$ cubing both sides, we get, ⇒ $k^3+\frac{1}{k^3}+3×k\times \frac{1}{k}(k+\frac{1}{k})=27$ ⇒ $k^3+\frac{1}{k^3}+3×3=27$ $\therefore k^3+\frac{1}{k^3}=27-9=18$ Hence, the correct answer is 18.
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