here is what i done
(1+i)^3=2*sqrt(2)( cos 3pi/4+ i sin 3pi/4)
(sqrt3 + i)^3=8 ( cos 3pi/6 +i sin 3pi/6)
so multiplying these two we get r1*r2= 16*sqrt(2)
and cos (theta1 + theta2)=cos (15pi/12)
and sin (theta1 + theta2)=sin (15pi/12)
so (1+i)^3(sqrt3 + i)^3=16*sqrt(2)(cos (15pi/12) +i sin (15pi/12))
= -16 - i 16
is that right?