• # question_answer If ${{(x+iy)}^{1/3}}=a+ib,$then $\frac{x}{a}+\frac{y}{b}$is equal to [IT 1982; Karnataka CET 2000] A) $4({{a}^{2}}+{{b}^{2}})$ B) $4({{a}^{2}}-{{b}^{2}})$ C) $4({{b}^{2}}-{{a}^{2}})$ D) None of these

${{(x+iy)}^{1/3}}=a+ib$Þ$(x+iy)={{(a+ib)}^{3}}$ $={{a}^{3}}+3{{a}^{2}}.ib+3a.{{(ib)}^{2}}+{{(ib)}^{3}}$ $={{a}^{3}}-3a{{b}^{2}}+i(3{{a}^{2}}b-{{b}^{3}})$ Equating real and imaginary parts, we get $\frac{x}{a}={{a}^{2}}-3{{b}^{2}}$and $\frac{y}{b}=3{{a}^{2}}-{{b}^{2}}$ $\therefore$  $\frac{x}{a}+\frac{y}{b}=4({{a}^{2}}-{{b}^{2}})$