Algebra Examples

$2 b^{3} + 5 = 2 {(3)}^{3} + 5$

Step 1

Simplify $2 \cdot 3^{3} + 5$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

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

$2 b^{3} + 5 = 2 \cdot 27 + 5$

Step 1.1.2

Multiply $2$ by $27$ .

$2 b^{3} + 5 = 54 + 5$

Step 1.2

Add $54$ and $5$ .

$2 b^{3} + 5 = 59$

Step 2

Move all terms not containing $b$ to the right side of the equation.

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Step 2.1

Subtract $5$ from both sides of the equation.

$2 b^{3} = 59 - 5$

Step 2.2

Subtract $5$ from $59$ .

$2 b^{3} = 54$

Step 3

Subtract $54$ from both sides of the equation.

$2 b^{3} - 54 = 0$

Step 4

Factor the left side of the equation.

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Step 4.1

Factor $2$ out of $2 b^{3} - 54$ .

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Step 4.1.1

Factor $2$ out of $2 b^{3}$ .

$2 (b^{3}) - 54 = 0$

Step 4.1.2

Factor $2$ out of $- 54$ .

$2 b^{3} + 2 \cdot - 27 = 0$

Step 4.1.3

Factor $2$ out of $2 b^{3} + 2 \cdot - 27$ .

$2 (b^{3} - 27) = 0$

Step 4.2

Rewrite $27$ as $3^{3}$ .

$2 (b^{3} - 3^{3}) = 0$

Step 4.3

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

$2 ((b - 3) (b^{2} + b \cdot 3 + 3^{2})) = 0$

Step 4.4

Factor.

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Step 4.4.1

Simplify.

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Step 4.4.1.1

Move $3$ to the left of $b$ .

$2 ((b - 3) (b^{2} + 3 b + 3^{2})) = 0$

Step 4.4.1.2

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

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

Step 4.4.2

Remove unnecessary parentheses.

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

Step 5

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

$b - 3 = 0$

$b^{2} + 3 b + 9 = 0$

Step 6

Set $b - 3$ equal to $0$ and solve for $b$ .

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Step 6.1

Set $b - 3$ equal to $0$ .

$b - 3 = 0$

Step 6.2

Add $3$ to both sides of the equation.

$b = 3$

Step 7

Set $b^{2} + 3 b + 9$ equal to $0$ and solve for $b$ .

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Step 7.1

Set $b^{2} + 3 b + 9$ equal to $0$ .

$b^{2} + 3 b + 9 = 0$

Step 7.2

Solve $b^{2} + 3 b + 9 = 0$ for $b$ .

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Step 7.2.1

Use the quadratic formula to find the solutions.

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

Step 7.2.2

Substitute the values $a = 1$ , $b = 3$ , and $c = 9$ into the quadratic formula and solve for $b$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot 9)}}{2 \cdot 1}$

Step 7.2.3

Simplify.

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Step 7.2.3.1

Simplify the numerator.

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Step 7.2.3.1.1

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

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

Step 7.2.3.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 7.2.3.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.3.1.2.2

Multiply $- 4$ by $9$ .

$b = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Step 7.2.3.1.3

Subtract $36$ from $9$ .

$b = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Step 7.2.3.1.4

Rewrite $- 27$ as $- 1 (27)$ .

$b = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Step 7.2.3.1.5

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$b = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.3.1.6

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

$b = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.3.1.7

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 7.2.3.1.7.1

Factor $9$ out of $27$ .

$b = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Step 7.2.3.1.7.2

Rewrite $9$ as $3^{2}$ .

$b = \frac{- 3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Step 7.2.3.1.8

Pull terms out from under the radical.

$b = \frac{- 3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Step 7.2.3.1.9

Move $3$ to the left of $i$ .

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

Step 7.2.3.2

Multiply $2$ by $1$ .

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

Step 7.2.4

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 7.2.4.1

Simplify the numerator.

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Step 7.2.4.1.1

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

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

Step 7.2.4.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 7.2.4.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.4.1.2.2

Multiply $- 4$ by $9$ .

$b = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Step 7.2.4.1.3

Subtract $36$ from $9$ .

$b = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Step 7.2.4.1.4

Rewrite $- 27$ as $- 1 (27)$ .

$b = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Step 7.2.4.1.5

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$b = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.4.1.6

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

$b = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.4.1.7

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 7.2.4.1.7.1

Factor $9$ out of $27$ .

$b = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Step 7.2.4.1.7.2

Rewrite $9$ as $3^{2}$ .

$b = \frac{- 3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Step 7.2.4.1.8

Pull terms out from under the radical.

$b = \frac{- 3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Step 7.2.4.1.9

Move $3$ to the left of $i$ .

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

Step 7.2.4.2

Multiply $2$ by $1$ .

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

Step 7.2.4.3

Change the $\pm$ to $+$ .

$b = \frac{- 3 + 3 i \sqrt{3}}{2}$

Step 7.2.4.4

Rewrite $- 3$ as $- 1 (3)$ .

$b = \frac{- 1 \cdot 3 + 3 i \sqrt{3}}{2}$

Step 7.2.4.5

Factor $- 1$ out of $3 i \sqrt{3}$ .

$b = \frac{- 1 \cdot 3 - (- 3 i \sqrt{3})}{2}$

Step 7.2.4.6

Factor $- 1$ out of $- 1 (3) - (- 3 i \sqrt{3})$ .

$b = \frac{- 1 (3 - 3 i \sqrt{3})}{2}$

Step 7.2.4.7

Move the negative in front of the fraction.

$b = - \frac{3 - 3 i \sqrt{3}}{2}$

Step 7.2.5

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 7.2.5.1

Simplify the numerator.

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Step 7.2.5.1.1

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

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

Step 7.2.5.1.2

Multiply $- 4 \cdot 1 \cdot 9$ .

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Step 7.2.5.1.2.1

Multiply $- 4$ by $1$ .

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

Step 7.2.5.1.2.2

Multiply $- 4$ by $9$ .

$b = \frac{- 3 \pm \sqrt{9 - 36}}{2 \cdot 1}$

Step 7.2.5.1.3

Subtract $36$ from $9$ .

$b = \frac{- 3 \pm \sqrt{- 27}}{2 \cdot 1}$

Step 7.2.5.1.4

Rewrite $- 27$ as $- 1 (27)$ .

$b = \frac{- 3 \pm \sqrt{- 1 \cdot 27}}{2 \cdot 1}$

Step 7.2.5.1.5

Rewrite $\sqrt{- 1 (27)}$ as $\sqrt{- 1} \cdot \sqrt{27}$ .

$b = \frac{- 3 \pm \sqrt{- 1} \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.5.1.6

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

$b = \frac{- 3 \pm i \cdot \sqrt{27}}{2 \cdot 1}$

Step 7.2.5.1.7

Rewrite $27$ as $3^{2} \cdot 3$ .

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Step 7.2.5.1.7.1

Factor $9$ out of $27$ .

$b = \frac{- 3 \pm i \cdot \sqrt{9 (3)}}{2 \cdot 1}$

Step 7.2.5.1.7.2

Rewrite $9$ as $3^{2}$ .

$b = \frac{- 3 \pm i \cdot \sqrt{3^{2} \cdot 3}}{2 \cdot 1}$

Step 7.2.5.1.8

Pull terms out from under the radical.

$b = \frac{- 3 \pm i \cdot (3 \sqrt{3})}{2 \cdot 1}$

Step 7.2.5.1.9

Move $3$ to the left of $i$ .

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

Step 7.2.5.2

Multiply $2$ by $1$ .

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

Step 7.2.5.3

Change the $\pm$ to $-$ .

$b = \frac{- 3 - 3 i \sqrt{3}}{2}$

Step 7.2.5.4

Rewrite $- 3$ as $- 1 (3)$ .

$b = \frac{- 1 \cdot 3 - 3 i \sqrt{3}}{2}$

Step 7.2.5.5

Factor $- 1$ out of $- 3 i \sqrt{3}$ .

$b = \frac{- 1 \cdot 3 - (3 i \sqrt{3})}{2}$

Step 7.2.5.6

Factor $- 1$ out of $- 1 (3) - (3 i \sqrt{3})$ .

$b = \frac{- 1 (3 + 3 i \sqrt{3})}{2}$

Step 7.2.5.7

Move the negative in front of the fraction.

$b = - \frac{3 + 3 i \sqrt{3}}{2}$

Step 7.2.6

The final answer is the combination of both solutions.

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

Step 8

The final solution is all the values that make $2 (b - 3) (b^{2} + 3 b + 9) = 0$ true.

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