Question

8y³ $-$ $\frac{125}{y^3}$

Answer 1

$$\begin{gathered} 8y^3-\frac{125}{y^3} \\ ={\left(2y\right)}^3-{\left(\frac{5}{y}\right)}^3 \\ =\left(2y-\frac{5}{y}\right)\left[{\left(2y\right)}^2+2y\times \frac{5}{y}+{\left({\displaystyle \frac{5}{y}}\right)}^2\right] \\ =\left(2y-\frac{5}{y}\right)\left(4y^2+10+\frac{25}{y^2}\right) \end{gathered}$$