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# Solve the cubic equation:

## $$x^3-3x^2+x+1=0$$

Since the discriminant $$\Delta <0$$, the cubic equation has three distinct real roots.

$$\Delta=-0.2962962962963$$

$$\begin{cases} x_1=1 \\ x_2=1+\sqrt{2} \\ x_3=1-\sqrt{2} \end{cases}$$

Detailed Steps on Solution

A cubic equation has at least one real root. If the coefficient of leading term is 1, one of solutions could be a factor of the constant term.

## 1. Factorization Method

Find all possible factors for constant

$$1$$

$$-1$$

Substitute the factors to the function $$f(x) = x³ - 3x² + x + 1$$ and find the one that makes $$f(x) = 0$$.

According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ - 3x² + x + 1$$ has a factor $$x-n$$, that is, $$x=n$$ is a root of the equation.

Fortunetely, one of the numbers is found to make the equation equal.

$$f(1) = 1³ - 3*1² + *1 + 1 = 0$$

then we get the first root

$$x_1 = 1$$

And the cubic equation can be factored as

$$(x -1)(ax^2+bx+c) = 0$$

Next we can use either long division or synthetic division to determine the expression of trinomial

### Long division

Divide the polynomial $$x³ - 3x² + x + 1$$ by $$x - 1$$

 x² -2x -1 x - 1 x³ -3x² +x +1 x³ -x² -2x² +x -2x² +2x -x +1 -x +1 0

Now we get another factor of the cubic equation $$x² - 2x - 1$$

## Solve the quadratic equation: $$x² - 2x - 1 = 0$$

Given $$a =1, b=-2, c=-1$$,

Use the root solution formula for a quadratic equation, the roots of the equation are given as

\begin{aligned} \\t&=\dfrac{-b\pm\sqrt{b^2-4ac} }{2a}\\ & =\dfrac{-(-2)\pm\sqrt{(-2)^2-4\cdot 1\cdot (-1)}}{2 \cdot 1}\\ & =\dfrac{2\pm2\sqrt{2}}{2}\\ & =1\pm\sqrt{2}\\ \end{aligned}

Since the discriminat is greater than zero, we get another two real roots:

That is,

$$\begin{cases} t_2 =1+\sqrt{2} \\ t_3=1-\sqrt{2} \end{cases}$$

Another method to find the roots of a general cubic equation is simplifing it to a depressed form. This method is applicable to all cases, especially to those difficult to find factors.

## 2. Convert to depressed cubic equation

The idea is to convert general form of cubic equation

$$ax^3+bx^2+cx+d = 0$$

to the form without quadratic term.

$$t^3+pt+q = 0$$

By substituting $$x$$ with $$t - \dfrac{b}{3a}$$, the general cubic equation could be transformed to

$$t^3+\dfrac{3ac-b^2}{3a^2}t+\dfrac{2b^3-9abc+27a^2d}{27a^3} = 0$$

Compare with the depressed cubic equation. Then,

$$p = \dfrac{3ac-b^2}{3a^2}$$

$$q = \dfrac{2b^3-9abc+27a^2d}{27a^3}$$

Substitute the values of coefficients, $$p, q$$ is obtained as

$$p = \dfrac{3\cdot 1\cdot 1-(-3)^2}{3\cdot 1^2}=-2$$

$$q = \dfrac{2\cdot (-3)^3-9\cdot1\cdot (-3)\cdot 1+27\cdot 1^2\cdot1}{27\cdot 1^3}=0$$

### Use the substitution to transform

Let $$p$$ and $$q$$ being the coefficient of the linean and constant terms, the depressed cubic equation is expressed as.

$$t^3 +pt+q=0$$

Let $$x=t+1$$

The cubic equation $$x³ - 3x² + x + 1=0$$ is transformed to

$$t^3 -2t=0$$

For this equation, we have $$p=-2$$ and $$q = 0$$

## Factor out common factor

Since each term contains a common factor, we can factor out the common factor $$x$$ out of each term.

$$(t² - 2)t = 0$$

In order for the equation to hold true, either of the factors is equal to 0. Then, we get the first root,

$$t_1 = 0$$

$$t² - 2 = 0$$

And then, the problem turns to solving a quadratic equation.

## Solve the quadratic equation: $$x² - 2 = 0$$

$$t^2 = 2$$

Solving for $$t$$.

Then,

\begin{aligned} \\ t&=\pm \sqrt{2}\\ &=\pm \sqrt{2}\\ \end{aligned}

That is,

$$\begin{cases} t_2 =\sqrt{2} \\ t_3=-\sqrt{2} \end{cases}$$

Since $$x = t - \dfrac{b}{3a}$$, substituting the values of $$t$$, $$a$$ and $$b$$ gives

$$x_1 = t_1+1=1$$

$$x_2 = t_2+1=1+\sqrt{2}$$

$$x_3 = t_3+1=1-\sqrt{2}$$

## 3. Summary

In summary, we have tried the method of factorization, cubic root formula to explore the solutions of the equation. The cubic equation $$x³ - 3x² + x + 1=0$$ is found to have three real roots . Exact values and approximations are given below.

$$\begin{cases} x_1=1 \\ x_2=1+\sqrt{2} \\ x_3=1-\sqrt{2} \end{cases}$$

Convert to decimals,

$$\begin{cases} x_1=1 \\ x_2=2.4142135623731 \\ x_3=-0.4142135623731 \end{cases}$$

Using the method of factorization, the roots are derived to the following forms

$$\begin{cases} x_1=1 \\ x_2=1-\sqrt{2} \\ x_3=1+\sqrt{2} \end{cases}$$

## 4. Graph for the function $$f(x) = x³ - 3x² + x + 1$$

Since the discriminat is less than zero, the curve of the cubic function $$f(x) = x³ - 3x² + x + 1$$ has 3 intersection points with horizontal axis.

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