Question

Find the value
${(2p-\frac{1}{2p})}^3$

Answer 1

here, ${(2p-\frac{1}{2p})}^3$

Using identity:- ${(a-b)}^3=a^3-b^3-3a^{2}b+3a{b}^2$

Therefore,

${(2p-\frac{1}{2p})}^3$

$={\left(2p\right)}^3-{\left(\frac{1}{2p}\right)}^3-3{\left(2p\right)}^2\left(\frac{1}{2p}\right)+3\left(2p\right){\left(\frac{1}{2p}\right)}^2$

$=8p^3-\frac{1}{8p^3}-3\left(4p^2\right)\left(\frac{1}{2p}\right)+3\left(2p\right)\left(\frac{1}{4p^2}\right)$

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

Hence ${(2p-\frac{1}{2p})}^3=8p^3-\frac{1}{8p^3}-6p+\frac{3}{2p}$