Solve the cubic equation:

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

Quick Answer

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 $$x + 1$$

+2x+1
x + 1+3x²+3x+1
+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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