Arun
Last Activity: 6 Years ago
Dear student
(x + iy) = [ a ^3 + (ib)^3 + 3abi (a + ib) ]
= a^3 - ib^3 + 3a^2bi - 3ab^2
= (a^3 - 3ab^2) + i (3a^2b - b^3 )
=> (x + iy) = a (a^2 - 3b^2 ) + ib (3a^2 - b^2 )
comparing the value of x and y both sides
x = a(a^2 - 3b^2) , y = b (3a^2 - b^2)
therefore (x/a) = (a^2 - 3b^2) and (y/b) = (3a^2 - b^2)
(x/a) + (y/b) = 4a^2 - 4b^2 = 4 (a^2 - b^2)
hence proved
Regards
Arun (askIITians forum expert)