A) 2
B) 4
C) 6
D) 1
Correct Answer: B
Solution :
\[{{(x+iy)}^{1/3}}=a-ib\] \[x+iy={{(a-ib)}^{3}}=({{a}^{3}}-3a{{b}^{2}})+i({{b}^{3}}-3{{a}^{2}}b)\] Þ \[x={{a}^{3}}-3a{{b}^{2}},\,y={{b}^{3}}-3{{a}^{2}}b\] Þ \[\frac{x}{a}={{a}^{2}}-3{{b}^{2}},\,\frac{y}{b}={{b}^{2}}-3{{a}^{2}}\] \[\therefore \] \[\frac{x}{a}-\frac{y}{b}={{a}^{2}}-3{{b}^{2}}-{{b}^{2}}+3{{a}^{2}}\] \[\frac{x}{a}-\frac{y}{b}=4({{a}^{2}}-{{b}^{2}})=k({{a}^{2}}-{{b}^{2}})\] \[\therefore \] \[k=4\].You need to login to perform this action.
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