If `f(x)` and `g(x)` are two polynomials such that the polynomial `h(x)=xf(x^3)+x^2g(x^6)` is divisible by `x^2+x+1,` then `f(1)=g(1)` (b) `f(1)=1g(1)` `h(1)=0` (d) all of these

Options

A. f(1)+g(1)=1
B. f(1)=-g(1)
C. f(1)=g(1) ` ne ` 0
D. `f(1)= pm g(1)`

Answer

To solve the problem, we need to analyze the polynomial \( h(x) = x f(x^3) + x^2 g(x^6) \) and determine the conditions under which it is divisible by \( x^2 + x + 1 \).

### Step 1: Identify the roots of the divisor
The polynomial \( x^2 + x + 1 \) has roots given by the cube roots of unity, specifically \( \omega \) and \( \omega^2 \), where:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2} i \]
and
\[ \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2} i. \]

### Step 2: Evaluate \( h(\omega) \) and \( h(\omega^2) \)
Since \( h(x) \) is divisible by \( x^2 + x + 1 \), it must hold that:
\[ h(\omega) = 0 \quad \text{and} \quad h(\omega^2) = 0. \]

Substituting \( \omega \) into \( h(x) \):
\[ h(\omega) = \omega f(\omega^3) + \omega^2 g(\omega^6). \]
Since \( \omega^3 = 1 \) and \( \omega^6 = 1 \), we can simplify this to:
\[ h(\omega) = \omega f(1) + \omega^2 g(1) = 0. \]

Similarly, substituting \( \omega^2 \):
\[ h(\omega^2) = \omega^2 f((\omega^2)^3) + \omega g((\omega^2)^6) = \omega^2 f(1) + \omega g(1) = 0. \]

### Step 3: Set up the equations
From the evaluations, we have two equations:
1. \( \omega f(1) + \omega^2 g(1) = 0 \) (Equation 1)
2. \( \omega^2 f(1) + \omega g(1) = 0 \) (Equation 2)

### Step 4: Solve the equations
We can solve these equations simultaneously. From Equation 1, we can express \( g(1) \) in terms of \( f(1) \):
\[ g(1) = -\frac{\omega}{\omega^2} f(1) = -\omega^3 f(1) = -f(1). \]

From Equation 2, we can express \( f(1) \) in terms of \( g(1) \):
\[ f(1) = -\frac{\omega^2}{\omega} g(1) = -\omega g(1). \]

### Step 5: Combine the results
Substituting \( g(1) = -f(1) \) into \( f(1) = -\omega g(1) \):
\[ f(1) = -\omega (-f(1)) = \omega f(1). \]
This implies:
\[ f(1)(1 - \omega) = 0. \]
Since \( 1 - \omega \neq 0 \), we conclude:
\[ f(1) = 0 \quad \text{and thus} \quad g(1) = 0. \]

### Conclusion
The conditions imply that \( f(1) = -g(1) \), which corresponds to option (b) \( f(1) = -g(1) \).