Question

If $p-\frac{1}{p}=4$, find the value of $p^4+\frac{1}{p^4}$.

• A. $16$
• B. $18$
• C. $324$
• D. $322$

Answer 1

The correct option is D $322$

Consider $p-\frac{1}{p}=4$, squaring both sides, we get:

${(p-\frac{1}{p})}^2=4^2\Rightarrow p^2+{\left(\frac{1}{p}\right)}^2-(2\times p\times \frac{1}{p})=16(\because (a-b{)}^2=a^2+b^2-2ab)\Rightarrow p^2+\frac{1}{p^2}-2=16\Rightarrow p^2+\frac{1}{p^2}=16+2\Rightarrow p^2+\frac{1}{p^2}=18........(1)$

Now again squaring both sides of equation 1, we have:

${(p^2+\frac{1}{p^2})}^2={18}^2\Rightarrow {\left(p^2\right)}^2+{\left(\frac{1}{p^2}\right)}^2+(2\times p^2\times \frac{1}{p^2})=324(\because (a+b{)}^2=a^2+b^2+2ab)\Rightarrow p^4+\frac{1}{p^4}+2=324\Rightarrow p^4+\frac{1}{p^4}=324-2\Rightarrow p^4+\frac{1}{p^4}=322$

Hence, $p^4+\frac{1}{p^4}=322$.