Solve by Factoring 27x^3+27x^2+9x+1=0

$27 x^{3} + 27 x^{2} + 9 x + 1 = 0$

Factor the left side of the equation.

Regroup terms.

$27 x^{3} + 1 + 27 x^{2} + 9 x = 0$

Rewrite $27 x^{3}$ as ${(3 x)}^{3}$ .

${(3 x)}^{3} + 1 + 27 x^{2} + 9 x = 0$

Rewrite $1$ as $1^{3}$ .

${(3 x)}^{3} + 1^{3} + 27 x^{2} + 9 x = 0$

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = 3 x$ and $b = 1$ .

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

Simplify.

Apply the product rule to $3 x$ .

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

Raise $3$ to the power of $2$ .

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

Multiply $3$ by $- 1$ .

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

Multiply $- 3$ by $1$ .

$(3 x + 1) (9 x^{2} - 3 x + 1^{2}) + 27 x^{2} + 9 x = 0$

One to any power is one.

$(3 x + 1) (9 x^{2} - 3 x + 1) + 27 x^{2} + 9 x = 0$

Factor $9 x$ out of $27 x^{2} + 9 x$ .

Factor $9 x$ out of $27 x^{2}$ .

$(3 x + 1) (9 x^{2} - 3 x + 1) + 9 x (3 x) + 9 x = 0$

Factor $9 x$ out of $9 x$ .

$(3 x + 1) (9 x^{2} - 3 x + 1) + 9 x (3 x) + 9 x (1) = 0$

Factor $9 x$ out of $9 x (3 x) + 9 x (1)$ .

$(3 x + 1) (9 x^{2} - 3 x + 1) + 9 x (3 x + 1) = 0$

Factor $3 x + 1$ out of $(3 x + 1) (9 x^{2} - 3 x + 1) + 9 x (3 x + 1)$ .

Factor $3 x + 1$ out of $9 x (3 x + 1)$ .

$(3 x + 1) (9 x^{2} - 3 x + 1) + (3 x + 1) (9 x) = 0$

Factor $3 x + 1$ out of $(3 x + 1) (9 x^{2} - 3 x + 1) + (3 x + 1) (9 x)$ .

$(3 x + 1) (9 x^{2} - 3 x + 1 + 9 x) = 0$

Add $- 3 x$ and $9 x$ .

$(3 x + 1) (9 x^{2} + 6 x + 1) = 0$

Factor using the perfect square rule.

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

$(3 x + 1) ({(3 x)}^{2} + 6 x + 1) = 0$

Rewrite $1$ as $1^{2}$ .

$(3 x + 1) ({(3 x)}^{2} + 6 x + 1^{2}) = 0$

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$6 x = 2 \cdot (3 x) \cdot 1$

Rewrite the polynomial.

$(3 x + 1) ({(3 x)}^{2} + 2 \cdot (3 x) \cdot 1 + 1^{2}) = 0$

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = 3 x$ and $b = 1$ .

$(3 x + 1) {(3 x + 1)}^{2} = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$3 x + 1 = 0$

${(3 x + 1)}^{2} = 0$

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

The final solution is all the values that make $(3 x + 1) {(3 x + 1)}^{2} = 0$ true.

$x = - \frac{1}{3}$

Solve by Factoring 27x^3+27x^2+9x+1=0

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