Let P ( X ) P(x) P ( X ) Be A Polynomial With Real Coefficients Such That: ( X 2 + X + 1 ) P ( X − 1 ) = ( X 2 − X + 1 ) P ( X ) ∀ X ∈ R (x^2 + X + 1)P(x - 1) = (x^2 - X + 1)P(x) \quad \forall\ X \in \mathbb{R} ( X 2 + X + 1 ) P ( X − 1 ) = ( X 2 − X + 1 ) P ( X ) ∀ X ∈ R

$$$$

Introduction

In this article, we will be discussing a polynomial $P(x)$ with real coefficients that satisfies the equation $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$. We will also be given the additional information that $P(1) = 3$. Our goal is to evaluate the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$.

Understanding the Given Equation

The given equation $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$ can be rewritten as:

$$(x^2 + x + 1)P(x - 1) - (x^2 - x + 1)P(x) = 0$$

This equation can be further simplified by substituting $y = x - 1$, which gives us:

$$(y^2 + y + 2)P(y) - (y^2 + y + 1)P(y + 1) = 0$$

Properties of the Polynomial $P(x)$

From the given equation, we can see that the polynomial $P(x)$ satisfies the following properties:

• $P(x)$ has real coefficients.
• $P(x)$ satisfies the equation $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$.
• $P(1) = 3$.

Evaluating the Integral

To evaluate the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$, we can start by substituting $y = x - 1$, which gives us:

$$\int_0^1 \frac{P(x)}{x^2 + x + 1} dx = \int_{-1}^0 \frac{P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy$$

Using the Properties of $P(x)$

Using the properties of $P(x)$, we can rewrite the integral as:

$$\int_{-1}^0 \frac{P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy = \int_{-1}^0 \frac{(y^2 + y + 2)P(y) - (y^2 + y + 1)P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy$$

Simplifying the Integral

Simplifying the integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y) - (y^2 + y + 1)P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy = \int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{(y + 1)^2 + (y + 1) + 1} dy - \int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy$$

Evaluating the First Integral

Evaluating the first integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{(y + 1)^2 + (y + 1) + 1} dy = \int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{y^2 + 2y + 2} dy$$

Simplifying the First Integral

Simplifying the first integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{y^2 + 2y + 2} dy = \int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{(y + 1)^2 + 1} dy$$

Evaluating the Second Integral

Evaluating the second integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy = \int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{y^2 + 2y + 2} dy$$

Simplifying the Second Integral

Simplifying the second integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{y^2 + 2y + 2} dy = \int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{(y + 1)^2 + 1} dy$$

Combining the Integrals

Combining the two integrals, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{(y + 1)^2 + 1} dy - \int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{(y + 1)^2 + 1} dy$$

Evaluating the Combined Integral

Evaluating the combined integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y)}{(y + 1)^2 + 1} dy - \int_{-1}^0 \frac{(y^2 + y + 1)P(y + 1)}{(y + 1)^2 + 1} dy = \int_{-1}^0 \frac{(y^2 + y + 2)P(y) - (y^2 + y + 1)P(y + 1)}{(y + 1)^2 + 1} dy$$

Simplifying the Combined Integral

Simplifying the combined integral, we get:

$$\int_{-1}^0 \frac{(y^2 + y + 2)P(y) - (y^2 + y + 1)P(y + 1)}{(y + 1)^2 + 1} dy = \int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy$$

Evaluating the Final Integral

Evaluating the final integral, we get:

$$\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy = \int_{-1}^0 \frac{P(y)}{y^2 + 2y + 2} dy$$

Simplifying the Final Integral

Simplifying the final integral, we get:

$$\int_{-1}^0 \frac{P(y)}{y^2 + 2y + 2} dy = \int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy$$

Conclusion

In conclusion, we have evaluated the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$ using the properties of the polynomial $P(x)$ and the given equation $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$. We have shown that the integral can be evaluated as $\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy$.

Final Answer

The final answer is $\boxed{\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy}$.

Introduction

In our previous article, we discussed a polynomial $P(x)$ with real coefficients that satisfies the equation $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$. We also evaluated the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$ using the properties of the polynomial $P(x)$ and the given equation.

Q&A

Q: What is the given equation that the polynomial $P(x)$ satisfies?

A: The given equation is $(x^2 + x + 1)P(x - 1) = (x^2 - x + 1)P(x) \quad \forall\ x \in \mathbb{R}$.

Q: What are the properties of the polynomial $P(x)$?

A: The polynomial $P(x)$ has real coefficients and satisfies the given equation. Additionally, we are given that $P(1) = 3$.

Q: How did we evaluate the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$?

A: We evaluated the integral by substituting $y = x - 1$, which gave us $\int_{-1}^0 \frac{P(y + 1)}{(y + 1)^2 + (y + 1) + 1} dy$. We then used the properties of the polynomial $P(x)$ to simplify the integral and evaluate it as $\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy$.

Q: What is the final answer to the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$?

A: The final answer to the integral is $\boxed{\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy}$.

Q: What are some common applications of the given equation and the properties of the polynomial $P(x)$?

A: The given equation and the properties of the polynomial $P(x)$ have applications in various fields, including algebra, analysis, and number theory. They can be used to study the properties of polynomials, solve equations, and evaluate integrals.

Q: How can we use the given equation and the properties of the polynomial $P(x)$ to solve other problems?

A: We can use the given equation and the properties of the polynomial $P(x)$ to solve other problems by applying the same techniques and methods used to evaluate the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$. We can also use the properties of the polynomial $P(x)$ to study its behavior, find its roots, and evaluate its derivatives.

Conclusion

In this Q&A article, we discussed the given equation and the properties of the polynomial $P(x)$, and evaluated the integral $\int_0^1 \frac{P(x)}{x^2 + x + 1} dx$. We also answered some common questions about the given equation and the properties of the polynomial $P(x)$. We hope that this article has been helpful in the given equation and the properties of the polynomial $P(x)$, and in applying them to solve other problems.

Final Answer

The final answer is $\boxed{\int_{-1}^0 \frac{P(y)}{(y + 1)^2 + 1} dy}$.