According to DeMoivre's theorem
#(r(costheta+isintheta))^n=r^n(cosntheta+isinntheta)#
Now let #sqrt5-4i=r(costheta+isintheta)#
hence #rcostheta=1# and #rsintheta=-sqrt3#
hence squaring and adding #r^2=((sqrt5)^2+(-4)^2)=5+16=21#
and #r=sqrt21#, #costheta=sqrt(5/21)# and #sintheta=-4/sqrt21#
Therefore using DeMoivre's theorem
#(sqrt5-4i)^3=(sqrt21)^3(cos3theta+isin3theta)#
As #cos3theta=4cos^3theta-3costheta=4(sqrt(5/21))^3-3sqrt(5/21)#
= #20/21sqrt(5/21)-3sqrt(5/21)=-43/21sqrt(5/21)#
and #sin3theta=3sintheta-4sin^3theta=3(-4/sqrt21)-4(-4/sqrt21)^3#
= #-12/sqrt21+256/(21sqrt21)=(-252+256)/(21sqrt21)=-4/(21sqrt21)#
Hence, #(sqrt5-4i)^3=(sqrt21)^3(-43/21sqrt(5/21)-i4/(21sqrt21))#
= #-43sqrt5-4i#
