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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