Question : If $\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)^2=9$, then what is the value of $\mathrm{k}^3+\frac{1}{\mathrm{k}^3} ?$
Option 1: 27
Option 2: 18
Option 3: 15
Option 4: 21
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Correct Answer: 18
Solution :
$\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)^2=9$
$⇒\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right) = 3$
$⇒\left(\mathrm{k}+\frac{1}{\mathrm{k}}\right)^3 = 3^3$
$⇒\mathrm{k}^3+ \frac{1}{\mathrm{k}^3}+3×\mathrm{k}×\frac{1}{\mathrm{k}}(\mathrm{k}+\frac{1}{\mathrm{k}})=27$
$⇒ \mathrm{k}^3 + \frac{1}{\mathrm{k}^3}=27- 3(\mathrm{k} + \frac{1}{\mathrm{k}} )$
$⇒\mathrm{k}^3 + \frac{1}{\mathrm{k}^3} = 27 - 3\times3$
$⇒\mathrm{k}^3 + \frac{1}{\mathrm{k}^3}=18$
Hence, the correct answer is 18.
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