Solve for x (x^2+3x+3)^(4/3)=1

${(x^{2} + 3 x + 3)}^{\frac{4}{3}} = 1$

Step 1

Raise each side of the equation to the power of $\frac{3}{4}$ to eliminate the fractional exponent on the left side.

${({(x^{2} + 3 x + 3)}^{\frac{4}{3}})}^{\frac{3}{4}} = \pm 1^{\frac{3}{4}}$

Step 2

Simplify the exponent.

Simplify the left side.

Simplify ${({(x^{2} + 3 x + 3)}^{\frac{4}{3}})}^{\frac{3}{4}}$ .

Multiply the exponents in ${({(x^{2} + 3 x + 3)}^{\frac{4}{3}})}^{\frac{3}{4}}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(x^{2} + 3 x + 3)}^{\frac{4}{3} \cdot \frac{3}{4}} = \pm 1^{\frac{3}{4}}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(x^{2} + 3 x + 3)}^{\frac{4}{3} \cdot \frac{3}{4}} = \pm 1^{\frac{3}{4}}$

Rewrite the expression.

${(x^{2} + 3 x + 3)}^{\frac{1}{3} \cdot 3} = \pm 1^{\frac{3}{4}}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(x^{2} + 3 x + 3)}^{\frac{1}{3} \cdot 3} = \pm 1^{\frac{3}{4}}$

Rewrite the expression.

${(x^{2} + 3 x + 3)}^{1} = \pm 1^{\frac{3}{4}}$

Simplify.

$x^{2} + 3 x + 3 = \pm 1^{\frac{3}{4}}$

Simplify the right side.

One to any power is one.

$x^{2} + 3 x + 3 = \pm 1$

Step 3

The complete solution is the result of both the positive and negative portions of the solution.

First, use the positive value of the $\pm$ to find the first solution.

$x^{2} + 3 x + 3 = 1$

Subtract $1$ from both sides of the equation.

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

Subtract $1$ from $3$ .

$x^{2} + 3 x + 2 = 0$

Factor $x^{2} + 3 x + 2$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $2$ and whose sum is $3$ .

$1, 2$

Write the factored form using these integers.

$(x + 1) (x + 2) = 0$

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

$x + 1 = 0$

$x + 2 = 0$

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Set $x + 2$ equal to $0$ and solve for $x$ .

Set $x + 2$ equal to $0$ .

$x + 2 = 0$

Subtract $2$ from both sides of the equation.

$x = - 2$

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

$x = - 1, - 2$

Next, use the negative value of the $\pm$ to find the second solution.

$x^{2} + 3 x + 3 = - 1$

Move all terms to the left side of the equation and simplify.

Add $1$ to both sides of the equation.

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

Add $3$ and $1$ .

$x^{2} + 3 x + 4 = 0$

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 1$ , $b = 3$ , and $c = 4$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 4)}}{2 \cdot 1}$

Simplify.

Simplify the numerator.

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

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot 4}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot 4$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 4}}{2 \cdot 1}$

Multiply $- 4$ by $4$ .

$x = \frac{- 3 \pm \sqrt{9 - 16}}{2 \cdot 1}$

Subtract $16$ from $9$ .

$x = \frac{- 3 \pm \sqrt{- 7}}{2 \cdot 1}$

Rewrite $- 7$ as $- 1 (7)$ .

$x = \frac{- 3 \pm \sqrt{- 1 \cdot 7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1 (7)}$ as $\sqrt{- 1} \cdot \sqrt{7}$ .

$x = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{7}}{2 \cdot 1}$

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \frac{- 3 \pm i \sqrt{7}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm i \sqrt{7}}{2}$

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{7}}{2}, - \frac{3 + i \sqrt{7}}{2}$

The complete solution is the result of both the positive and negative portions of the solution.

$x = - 1, - 2, - \frac{3 - i \sqrt{7}}{2}, - \frac{3 + i \sqrt{7}}{2}$

Solve for x (x^2+3x+3)^(4/3)=1

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