If `(x+iy)^(1//3)=a+ib, " then " (x)/(a)+(y)/(b)` equals

Options

A. `4(a^(2)-b^(2))`
B. `2(a^(2)-b^(2))`
C. `2(a^(2)+b^(2))`
D. None of these

Explanation

To solve the problem, we start with the equation given:

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

### Step 1: Cube both sides
To eliminate the cube root, we cube both sides of the equation:

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

### Step 2: Expand the right-hand side
Using the binomial expansion, we expand \((a + ib)^3\):

\[ (a + ib)^3 = a^3 + 3a^2(ib) + 3a(ib)^2 + (ib)^3 \]

Calculating each term:
- \( (ib)^2 = -b^2 \)
- \( (ib)^3 = -ib^3 \)

Thus, we have:

\[ = a^3 + 3a^2(ib) - 3ab^2 - ib^3 \]

Combining the real and imaginary parts gives us:

\[ = (a^3 - 3ab^2) + i(3a^2b - b^3) \]

### Step 3: Equate real and imaginary parts
Now we equate the real and imaginary parts from both sides:

1. Real part: \( x = a^3 - 3ab^2 \)
2. Imaginary part: \( y = 3a^2b - b^3 \)

### Step 4: Express \( \frac{x}{a} + \frac{y}{b} \)
We need to find \( \frac{x}{a} + \frac{y}{b} \):

\[ \frac{x}{a} = \frac{a^3 - 3ab^2}{a} = a^2 - 3b^2 \]

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

Adding these two results together:

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

### Step 5: Combine like terms
Combining the terms gives:

\[ = a^2 + 3a^2 - 3b^2 - b^2 = 4a^2 - 4b^2 \]

### Final Result
Thus, we have:

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