If `x, y, a, b` are real numbers such that `(x+iy)^(1//5)=a + ib and p = (x)/(a) - (y)/(b)`, then

Options

A. `a-b` is factor of p
B. `a+b` is a factor of p
C. `a+ib` is a factor of p
D. `a-ib` is a factor of p

Answer

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

\[ (x + iy)^{1/5} = a + ib \]

We can raise both sides to the power of 5:

\[ x + iy = (a + ib)^5 \]

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

\[ (a + ib)^5 = \sum_{k=0}^{5} \binom{5}{k} a^{5-k} (ib)^k \]

This expands to:

\[ = a^5 + 5a^4(ib) + 10a^3(ib)^2 + 10a^2(ib)^3 + 5a(ib)^4 + (ib)^5 \]

Now, substituting \(i^2 = -1\), \(i^3 = -i\), \(i^4 = 1\), and \(i^5 = i\):

\[ = a^5 + 5a^4(ib) + 10a^3(-b^2) + 10a^2(-ib^3) + 5a(b^4) + (ib^5) \]

This simplifies to:

\[ = a^5 - 10a^3b^2 + 5ab^4 + i(5a^4b - 10a^2b^3 + b^5) \]

Now, we can equate the real and imaginary parts:

1. Real part:
\[ x = a^5 - 10a^3b^2 + 5ab^4 \]

2. Imaginary part:
\[ y = 5a^4b - 10a^2b^3 + b^5 \]

Next, we need to find \(p\) defined as:

\[ p = \frac{x}{a} - \frac{y}{b} \]

Substituting the expressions for \(x\) and \(y\):

\[ p = \frac{a^5 - 10a^3b^2 + 5ab^4}{a} - \frac{5a^4b - 10a^2b^3 + b^5}{b} \]

This simplifies to:

\[ p = a^4 - 10a^2b^2 + 5b^3 - (5a^3 - 10ab^2 + \frac{b^5}{b}) \]

Combining the terms, we get:

\[ p = a^4 - 10a^2b^2 + 5b^3 - 5a^3 + 10ab^2 - b^4 \]

Now, we can factor \(p\). We notice that:

\[ p = 4(b^2 - a^2)(b^2 + a^2) \]

Thus, we can express \(p\) as:

\[ p = 4(b^2 - a^2)(b^2 + a^2) \]

Finally, we can factor \(b^2 - a^2\) as \((b - a)(b + a)\):

\[ p = 4(b - a)(b + a)(b^2 + a^2) \]