Question

If ${\left(p^2+q^2\right)}^3={\left(p^3+q^3\right)}^2$ and $pq\ne 0,$ then the value of $\frac{p}{q}+\frac{q}{p}$ is

Answer 1

Consider the given equation.

${(p^2+q^2)}^3={(p^3+q^3)}^2$

${(p^2+q^2)}^3={(p^3+q^3)}^2$

$p^6+q^6+3p^2{q}^2(p^2+q^2)=(p^6+q^6+2p^3{q}^3)$

$3p^4{q}^2+3p^2{q}^4=2p^3{q}^3$

$3p^2{q}^2(p^2+q^2)=2p^3{q}^3$

$p^2+q^2=\frac{2}{3}pq$

$\frac{p}{q}+\frac{q}{p}=\frac{2}{3}$

Hence, the value is $\frac{2}{3}$.