(x+1)(x^2+x+1)(x-1)(x^2-x+1)=7

3 min read Jun 16, 2024

Solving the Equation: (x+1)(x^2+x+1)(x-1)(x^2-x+1) = 7

This equation presents a unique challenge due to its complex factorization. Let's break down the process of finding its solutions:

Simplifying the Expression

Notice that the equation has a pattern of conjugate pairs:

(x + 1) and (x - 1) are conjugates.
(x² + x + 1) and (x² - x + 1) are also conjugates.

This is helpful because multiplying conjugate pairs simplifies the expression significantly.

Recall: (a + b)(a - b) = a² - b²

Applying this to our equation:

(x + 1)(x - 1) = x² - 1
(x² + x + 1)(x² - x + 1) = (x²)² - (x)² + 1 = x⁴ - x² + 1

Now our equation becomes: (x² - 1)(x⁴ - x² + 1) = 7

Expanding and Rearranging

Let's expand the left side and rearrange the equation to get a standard polynomial form:

x⁶ - x⁴ + x² - x⁴ + x² - 1 = 7
x⁶ - 2x⁴ + 2x² - 8 = 0

Solving the Equation

This sixth-degree polynomial equation doesn't have a simple analytical solution. Here's why:

No Easy Factorization: The equation doesn't readily factor into simpler expressions.
No Rational Root Theorem: The Rational Root Theorem doesn't guarantee any rational roots for this equation.

To find the solutions, we'll need to employ numerical methods:

Graphing: Plotting the equation y = x⁶ - 2x⁴ + 2x² - 8 will show the points where the graph intersects the x-axis. These intersection points represent the real solutions.
Numerical Solvers: Software tools like Wolfram Alpha, Mathematica, or online equation solvers can provide numerical approximations for the roots.

Conclusion

The equation (x+1)(x^2+x+1)(x-1)(x^2-x+1)=7 presents a complex challenge that requires numerical methods to solve. While we cannot find exact solutions analytically, we can use graphical and numerical techniques to approximate the real roots of the equation.