Question

If (x + iy)1/3 = a + ib, then \[\frac{x}{a} + \frac{y}{b} =\]

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Answer 1

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\[\left( x + iy \right)^\frac{1}{3} = a + ib\]

\[\text { Cubing on both the sides, we get }: \]

\[x + iy = \left( a + ib \right)^3 \]

\[ \Rightarrow x + iy = a^3 + \left( ib \right)^3 + 3 a^2 bi + 3a \left( ib \right)^2 \]

\[ \Rightarrow x + iy = a^3 + i^3 b^3 + 3 a^2 ib + 3 i^2 a b^2 \]

\[ \Rightarrow x + iy = a^3 - i b^3 + 3 a^2 ib - 3a b^2 ( \because i^2 = - 1, i^3 = - i)\]

\[ \Rightarrow x + iy = a^3 - 3a b^2 + i\left( - b^3 + 3 a^2 b \right)\]

\[ \therefore x = a^3 - 3a b^2 \text { and }y = 3 a^2 b - b^3 \]

\[or , \frac{x}{a} = a^2 - 3 b^2\text {  and } \frac{y}{b} = 3 a^2 - b^2 \]

\[ \Rightarrow \frac{x}{a} + \frac{y}{b} = a^2 - 3 b^2 + 3 a^2 - b^2 \]

\[ \Rightarrow \frac{x}{a} + \frac{y}{b} = 4 a^2 - 4 b^2\]