Question

If $p+\frac{1}{p}=3$ then find the value of $\frac{p^4+\frac{1}{p^2}}{p^2+3p+1}$.

Answer 1

Given,

$p+\frac{1}{p}=3$......(1).

Then,

$=\frac{p(p^3+\frac{1}{p^3})}{p(p+3+\frac{1}{p})}$.

$=\frac{(p^3+\frac{1}{p^3})}{(p+3+\frac{1}{p})}$

$=\frac{{(p+\frac{1}{p})}^3-3.p.\frac{1}{p}.(p+\frac{1}{p})}{(p+3+\frac{1}{p})}$.

$=\frac{3^3-3.3}{3+3}$ [ Using (1)]

$=\frac{18}{6}=3$.