Question

If $(3p-\frac{1}{3p})=5$ and $(3p+\frac{1}{3p})=8$, then find the value of $9p^2-\frac{1}{9p^2}$

Answer 1

It is given that $(3p-\frac{1}{3p})=5$ and $(3p+\frac{1}{3p})=8$. We consider the product of both the given terms as shown below:

$(3p-\frac{1}{3p})\times (3p+\frac{1}{3p})=5\times 8\Rightarrow {\left(3p\right)}^2-{\left(\frac{1}{3p}\right)}^2=40(\because (a-b\left)\right(a+b)=a^2-b^2)\Rightarrow 9p^2-\frac{1}{9p^2}=40$

Hence, $9p^2-\frac{1}{9p^2}=40$.