C-03-Q098mediumsingle_mcqIf m+1m=am + \dfrac{1}{m} = am+m1=a, m3+1m3=?m^3 + \dfrac{1}{m^3} = ?m3+m31=?aa3−3aa^3 - 3aa3−3aba3+3aa^3 + 3aa3ব্যাখ্যাBy the identity m3+1m3=(m+1m)3−3(m+1m)=a3−3am^3+\dfrac1{m^3}=\left(m+\dfrac1m\right)^3-3\left(m+\dfrac1m\right)=a^3-3am3+m.