Question : If $p^2+\frac{1}{p^2}=14$, then find the value of $\left(p^3+\frac{1}{p^3}\right)$.
Option 1: 56
Option 2: 60
Option 3: 48
Option 4: 52
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Correct Answer: 52
Solution : Given: $p^2+\frac{1}{p^2}=14$ we know that, $p^2+\frac{1}{p^2}=(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}$ $⇒(p+\frac{1}{p})^2-2\times p\times \frac{1}{p}=14$ $⇒(p+\frac{1}{p})^2=14+2$ $⇒(p+\frac{1}{p})^2=16$ $⇒(p+\frac{1}{p})=4$ Now cubing both sides, we get: $⇒(p+\frac{1}{p})^3=4^3$ $⇒p^3+\frac{1}{p^3}+3\times p\times \frac{1}{p}(p+ \frac{1}{p})=4^3$ $⇒p^3+\frac{1}{p^3}+3\times4=64$ $⇒p^3+\frac{1}{p^3}=64-12$ $⇒p^3+\frac{1}{p^3}=52$ Hence, the correct answer is 52.
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