Question

विस्तार करा: $\bigg \lgroup 2p-\frac{1}{2p} \bigg \rgroup^3$

Answer 1

heya!!!!

your answer

thanks

Answer 2

विस्तार सूत्र:

$( {x-y)}^{3} = {x}^{3} - {y}^{3} - 3 {x}^{2} y + 3x {y}^{2}$
विस्तार करा:$\bigg \lgroup 2p-\frac{1}{2p} \bigg \rgroup^3$

$x = 2p\\ \\ y= \frac{1}{2p}$
$\bigg( { 2p - \frac{1}{2p} \bigg)}^{3} = {(2p)}^{3} - \bigg({ \frac{1}{2p} \bigg)}^{3} - 3 {(2p)}^{2} \bigg( \frac{1}{2p}\bigg) + 3(2p) {\bigg( \frac{1}{2p}\bigg) }^{2} \\ \\ = 8p^{3}- \frac{1}{8 {p}^{3} } - 6p + \frac{3}{2p}$
विस्तार:

$\bigg \lgroup 2p-\frac{1}{2p} \bigg \rgroup^3 =8p^{3}- \frac{1}{8 {p}^{3} } - 6p + \frac{3}{2p}$