Let P(x) and Q(x) be two polynomials.Suppose that `f(x) = P(x^3) + x Q(x^3)` is divisible by `x^2 + x+1,` then (a) P(x) is divisible by (x-1),but Q(x) is not divisible by x -1 (b) Q(x) is divisible by (x-1), but P(x) is not divisible by x-1 (c) Both P(x) and Q(x) are divisible by x-1 (d) f(x) is divisible by x-1

Options

A. P(x) is divisible by (x-1),but Q(x) is not divisible by x -1
B. Q(x) is divisible by (x-1), but P(x) is not divisible by x-1
C. Both P(x) and Q(x) are divisible by x-1
D. f(x) is divisible by x-1

Explanation

To solve the problem, we need to analyze the polynomial \( f(x) = P(x^3) + x Q(x^3) \) and determine the implications of its divisibility by \( x^2 + x + 1 \).

### Step 1: Roots of the divisor
The polynomial \( x^2 + x + 1 \) has roots that can be found using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
For \( a = 1, b = 1, c = 1 \):
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot 1}}{2 \cdot 1} = \frac{-1 \pm \sqrt{1 - 4}}{2} = \frac{-1 \pm \sqrt{-3}}{2} = \frac{-1 \pm i\sqrt{3}}{2} \]
Let \( \omega = \frac{-1 + i\sqrt{3}}{2} \) and \( \omega^2 = \frac{-1 - i\sqrt{3}}{2} \). These are the cube roots of unity, satisfying \( \omega^3 = 1 \).

### Step 2: Evaluate \( f(\omega) \) and \( f(\omega^2) \)
Since \( f(x) \) is divisible by \( x^2 + x + 1 \), it must equal zero at its roots:
1. For \( \omega \):
\[ f(\omega) = P(\omega^3) + \omega Q(\omega^3) = P(1) + \omega Q(1) = 0 \]
2. For \( \omega^2 \):
\[ f(\omega^2) = P((\omega^2)^3) + \omega^2 Q((\omega^2)^3) = P(1) + \omega^2 Q(1) = 0 \]

### Step 3: Set up equations
From the evaluations:
1. \( P(1) + \omega Q(1) = 0 \) (Equation 1)
2. \( P(1) + \omega^2 Q(1) = 0 \) (Equation 2)

### Step 4: Solve the equations
Subtract Equation 1 from Equation 2:
\[ (P(1) + \omega^2 Q(1)) - (P(1) + \omega Q(1)) = 0 \]
This simplifies to:
\[ (\omega^2 - \omega) Q(1) = 0 \]
Since \( \omega^2 \neq \omega \), we conclude that \( Q(1) = 0 \).

Substituting \( Q(1) = 0 \) back into Equation 1:
\[ P(1) + \omega \cdot 0 = 0 \implies P(1) = 0 \]

### Step 5: Conclusion
Since both \( P(1) = 0 \) and \( Q(1) = 0 \), we conclude that both \( P(x) \) and \( Q(x) \) are divisible by \( x - 1 \).

### Final Answer
The correct option is:
(c) Both \( P(x) \) and \( Q(x) \) are divisible by \( x - 1 \).