Solving the Cubic Equation x3 - x2 - x 33 - 32 - 3

Let's explore a specific cubic equation and its solutions. We start with the given equation:

x3 - x2 - x 33 - 32 - 3

Step 1: Simplifying the Right-Hand Side

First, we simplify the right-hand side of the equation:

33  2732  9

Thus, the equation becomes:

x3 - x2 - x 27 - 9 - 3 15

Step 2: Rearranging into a Standard Polynomial Equation

We rearrange the equation to form a standard polynomial:

x3 - x2 - x - 15 0

Step 3: Solving for x

We will check if x 3 is a root, as it seems to be a solution to the problem.

Substitute x 3 into the left-hand side of the equation:

x3 - x2 - x - 15 (33 - 32 - 3) - 15

(27 - 9 - 3) - 15 0

Hence, x 3 is a solution. This confirms that x 3 is a valid root of the equation.

Generalizing the Solution

We can generalize the equation by replacing 3 with a parameter t:

x3 - x2 - x - t3 - t2 - t 0

In this more general form, we can factorize the equation and extract the roots:

For x 3, we can factorize the equation as follows:

x3 - x2 - x - 15  (x - 3)(x2   2x   5)  0

The quadratic factor can be solved using the quadratic formula:

x (frac{1}{2}(1 pm sqrt{3t^2 - 2t - 5}))

This quadratic factor has solutions given by:

x (frac{1}{2}(1 pm sqrt{1 - 3t - 2t}))

Depending on the value of t, the equation has:

Two distinct real solutions if -1 Two coincident solutions at x 1 if t -1 Two coincident solutions at x -(frac{1}{3}) if t (frac{5}{3}) A complex conjugate pair of solutions if t (frac{5}{3})

Conclusion

From the initial equation, the solution x 3 is confirmed. However, there may be other solutions depending on the value of the parameter t.

The complete solution for the equation x^3 - x^2 - x - t^3 - t^2 - t 0 involves:

x 3 (if t 3) The quadratic factor solutions (frac{1}{2}(1 pm sqrt{3t^2 - 2t - 5})) (if t (-1, frac{5}{3}))