Question

Please Please Please give me Right answer​

Answer 1

${m}^{3} + \frac{1}{64 {m}^{3} }$

$= {(m)}^{3} + { (\frac{1}{4m}) }^{3}$

$= (m + \frac{1}{4m} )( {m}^{2} - \frac{1}{4m} \times m + { (\frac{1}{4} )}^{2}$

$= (m + \frac{1}{4m} )( {m}^{2} - \frac{1}{4} + \frac{1}{16} )$