# Solve the cubic equation:

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

Since the discriminant $$\Delta=0$$, the cubic equation has three real roots, of which two are equal.

$$\Delta=0$$

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

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 $$f(x) = x³ + 3x² + 3x + 1$$ and find the one that makes $$f(x) = 0$$.

According to factor theorem, $$f(n) = 0$$, if and only if the polynomial $$x³ + 3x² + 3x + 1$$ 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.

$$f(-1) = (-1)³ + 3(-1)² + 3(-1) + 1 = 8$$

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 $$x³ + 3x² + 3x + 1$$ by $$x + 1$$

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

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

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

Since the polynomial on the left hand side is a perfect square, we can factor the expression directly using the identity of Perfect Square.

$$(t+1)^2=0$$

Then we get two equal roots.

That is,

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

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

$$p = \dfrac{3\cdot 1\cdot 3-3^2}{3\cdot 1^2}=0$$

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

### 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 $$x³ + 3x² + 3x + 1=0$$ is transformed to

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

$$t_1=t_2=t_3=0$$

Therefore,

$$x_1=x_2=x_3=-1$$

## 3. Use the sum of cubes identity

Evaluate the coefficients of the cubic equation.

$$x³ + 3x² + 3x + 1=0$$

Apply the sum of cubes identity

$$(x+1)^3=0$$

Then, we get three equal roots for the cubic equation.

$$x_1=x_2=x_3=-1$$

## 4. Summary

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

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

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

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

Since the discriminat is equal to zero, the curve of the cubic function $$f(x) = x³ + 3x² + 3x + 1$$ has one intersection point with the x-axis.

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