If `(x+i y)^(1//3)=a+i b ,x ,y ,a b in R.`Show that(i) `x/a+y/b=4(a^2-b^2)`(ii) `x/a-y/b=2(a^2+b^2)`

If `(x+i y)^(1//3)=a+i b ,x ,y ,a b in R.`Show that(i) `x/a+y/b=4(a^2-

1.	If `(x+i y)^(1//3)=a+i b ,x ,y ,a b in R.`Show that(i) `x/a+y/b=4(a^2-b^2)`(ii) `x/a-y/b=2(a^2+b^2)`
Answer» We have `(x+iy)^(1//3)=(a+ib)` `rArr" "(x+iy)=(a+ib)^(3)" "["on cubing both sides"]` `rArr" "(x+iy)=a^(3)+i^(3)b^(3)+3iab(a+ib)` `=a^(3)-ib^(3)+3a^(2)bi-3ab^(2)=(a^(3)-3ab^(2))+i(3a^(2)b-b^(3))` `rArr" "x = a^(3) - 3ab^(2) and y = 3a^(2)b-b^(3)" "["on equating real and imaginary parts separetely"]` `rArr" "(x)/(a)=(a^(2)-3b^(2))and(y)/(b)=(3a^(2)-b^(2))` `rArr" "((x)/(a)+(y)/(b))=4(a^(2)-b^(2))and ((x)/(a)-(y)/(b))= -2(a^(2)+b^(2))`. Hence, (i) `(x)/(a)+(y)/(b)=4(a^(2)-b^(2)) and (ii) (x)/(a)-(y)/(b) = -2(a^(2)+b^(2))`.