Basic Math Examples

$1 = 64 {(r)}^{3}$

Step 1

Rewrite the equation as $64 r^{3} = 1$ .

$64 r^{3} = 1$

Step 2

Subtract $1$ from both sides of the equation.

$64 r^{3} - 1 = 0$

Step 3

Factor the left side of the equation.

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Step 3.1

Rewrite $64 r^{3}$ as ${(4 r)}^{3}$ .

${(4 r)}^{3} - 1 = 0$

Step 3.2

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

${(4 r)}^{3} - 1^{3} = 0$

Step 3.3

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 $a = 4 r$ and $b = 1$ .

$(4 r - 1) ({(4 r)}^{2} + 4 r \cdot 1 + 1^{2}) = 0$

Step 3.4

Simplify.

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Step 3.4.1

Apply the product rule to $4 r$ .

$(4 r - 1) (4^{2} r^{2} + 4 r \cdot 1 + 1^{2}) = 0$

Step 3.4.2

Raise $4$ to the power of $2$ .

$(4 r - 1) (16 r^{2} + 4 r \cdot 1 + 1^{2}) = 0$

Step 3.4.3

Multiply $4$ by $1$ .

$(4 r - 1) (16 r^{2} + 4 r + 1^{2}) = 0$

Step 3.4.4

One to any power is one.

$(4 r - 1) (16 r^{2} + 4 r + 1) = 0$

Step 4

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

$4 r - 1 = 0$

$16 r^{2} + 4 r + 1 = 0$

Step 5

Set $4 r - 1$ equal to $0$ and solve for $r$ .

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Step 5.1

Set $4 r - 1$ equal to $0$ .

$4 r - 1 = 0$

Step 5.2

Solve $4 r - 1 = 0$ for $r$ .

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Step 5.2.1

Add $1$ to both sides of the equation.

$4 r = 1$

Step 5.2.2

Divide each term in $4 r = 1$ by $4$ and simplify.

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Step 5.2.2.1

Divide each term in $4 r = 1$ by $4$ .

$\frac{4 r}{4} = \frac{1}{4}$

Step 5.2.2.2

Simplify the left side.

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Step 5.2.2.2.1

Cancel the common factor of $4$ .

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Step 5.2.2.2.1.1

Cancel the common factor.

$\frac{4 r}{4} = \frac{1}{4}$

Step 5.2.2.2.1.2

Divide $r$ by $1$ .

$r = \frac{1}{4}$

Step 6

Set $16 r^{2} + 4 r + 1$ equal to $0$ and solve for $r$ .

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Step 6.1

Set $16 r^{2} + 4 r + 1$ equal to $0$ .

$16 r^{2} + 4 r + 1 = 0$

Step 6.2

Solve $16 r^{2} + 4 r + 1 = 0$ for $r$ .

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Step 6.2.1

Use the quadratic formula to find the solutions.

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

Step 6.2.2

Substitute the values $a = 16$ , $b = 4$ , and $c = 1$ into the quadratic formula and solve for $r$ .

$\frac{- 4 \pm \sqrt{4^{2} - 4 \cdot (16 \cdot 1)}}{2 \cdot 16}$

Step 6.2.3

Simplify.

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Step 6.2.3.1

Simplify the numerator.

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Step 6.2.3.1.1

Raise $4$ to the power of $2$ .

$r = \frac{- 4 \pm \sqrt{16 - 4 \cdot 16 \cdot 1}}{2 \cdot 16}$

Step 6.2.3.1.2

Multiply $- 4 \cdot 16 \cdot 1$ .

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Step 6.2.3.1.2.1

Multiply $- 4$ by $16$ .

$r = \frac{- 4 \pm \sqrt{16 - 64 \cdot 1}}{2 \cdot 16}$

Step 6.2.3.1.2.2

Multiply $- 64$ by $1$ .

$r = \frac{- 4 \pm \sqrt{16 - 64}}{2 \cdot 16}$

Step 6.2.3.1.3

Subtract $64$ from $16$ .

$r = \frac{- 4 \pm \sqrt{- 48}}{2 \cdot 16}$

Step 6.2.3.1.4

Rewrite $- 48$ as $- 1 (48)$ .

$r = \frac{- 4 \pm \sqrt{- 1 \cdot 48}}{2 \cdot 16}$

Step 6.2.3.1.5

Rewrite $\sqrt{- 1 (48)}$ as $\sqrt{- 1} \cdot \sqrt{48}$ .

$r = \frac{- 4 \pm \sqrt{- 1} \cdot \sqrt{48}}{2 \cdot 16}$

Step 6.2.3.1.6

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

$r = \frac{- 4 \pm i \cdot \sqrt{48}}{2 \cdot 16}$

Step 6.2.3.1.7

Rewrite $48$ as $4^{2} \cdot 3$ .

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Step 6.2.3.1.7.1

Factor $16$ out of $48$ .

$r = \frac{- 4 \pm i \cdot \sqrt{16 (3)}}{2 \cdot 16}$

Step 6.2.3.1.7.2

Rewrite $16$ as $4^{2}$ .

$r = \frac{- 4 \pm i \cdot \sqrt{4^{2} \cdot 3}}{2 \cdot 16}$

Step 6.2.3.1.8

Pull terms out from under the radical.

$r = \frac{- 4 \pm i \cdot (4 \sqrt{3})}{2 \cdot 16}$

Step 6.2.3.1.9

Move $4$ to the left of $i$ .

$r = \frac{- 4 \pm 4 i \sqrt{3}}{2 \cdot 16}$

Step 6.2.3.2

Multiply $2$ by $16$ .

$r = \frac{- 4 \pm 4 i \sqrt{3}}{32}$

Step 6.2.3.3

Simplify $\frac{- 4 \pm 4 i \sqrt{3}}{32}$ .

$r = \frac{- 1 \pm i \sqrt{3}}{8}$

Step 6.2.4

The final answer is the combination of both solutions.

$r = - \frac{1 - i \sqrt{3}}{8}, - \frac{1 + i \sqrt{3}}{8}$

Step 7

The final solution is all the values that make $(4 r - 1) (16 r^{2} + 4 r + 1) = 0$ true.

$r = \frac{1}{4}, - \frac{1 - i \sqrt{3}}{8}, - \frac{1 + i \sqrt{3}}{8}$