Question
Expand: $\left(2 p-\frac{1}{2 p}\right)^3$

Answer

$\begin{aligned} & \text { Here, } a=2 p \text { and } b=\frac{1}{2 p} \\ & \left(2 p-\frac{1}{2 p}\right)^3 \\ & =(2 p)^3-3(2 p)^2\left(\frac{1}{2 p}\right)+3(2 p)\left(\frac{1}{2 p}\right)^2-\left(\frac{1}{2 p}\right)^3 \\ & \quad \ldots\left[(a-b)^3=a^3-3 a^2 b+3 a b^2-b^3\right] \\ & =8 p^3-3(2 p)(2 p)\left(\frac{1}{2 p}\right)+3(2 p)\left(\frac{1}{2 p}\right)\left(\frac{1}{2 p}\right)-\frac{1}{8 p^3} \\ & =8 p^3-3(2 p)+3\left(\frac{1}{2 p}\right)-\frac{1}{8 p^3} \\ & \mathbf{8 p}^3-\mathbf{6 p}+\frac{\mathbf{3}}{2 p}-\frac{1}{8 p^3}\end{aligned}$

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