If `(x+iy)^(1//3)=a+ib`, where x,y,a,b`inR` show that
`(x)/(a)-(y)/(b)=-2(a^(2)+b^(2))`.

Answer

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

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

We will cube both sides to eliminate the cube root:

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

Next, we expand the right-hand side using the binomial expansion:

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

Calculating each term:

1. \( (ib)^3 = i^3b^3 = -ib^3 \)
2. \( 3a(ib)^2 = 3a(i^2b^2) = 3a(-b^2) = -3ab^2 \)
3. \( 3a^2(ib) = 3a^2ib \)

Putting it all together, we have:

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

Now, we can equate the real and imaginary parts:

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

Next, we need to find the expression \( \frac{x}{a} - \frac{y}{b} \):

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

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

Now, substituting these into the expression:

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

Simplifying this gives:

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

Thus, we can write:

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

This completes the proof.