Precalculus Examples

${(8 x - 7)}^{\frac{1}{3}} = x^{\frac{2}{3}}$

Step 1

Eliminate the fractional exponents by multiplying both exponents by the LCD.

${({(8 x - 7)}^{\frac{1}{3}})}^{3} = {(x^{\frac{2}{3}})}^{3}$

Step 2

Simplify ${({(8 x - 7)}^{\frac{1}{3}})}^{3}$ .

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Step 2.1

Multiply the exponents in ${({(8 x - 7)}^{\frac{1}{3}})}^{3}$ .

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Step 2.1.1

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

${(8 x - 7)}^{\frac{1}{3} \cdot 3} = {(x^{\frac{2}{3}})}^{3}$

Step 2.1.2

Cancel the common factor of $3$ .

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Step 2.1.2.1

Cancel the common factor.

${(8 x - 7)}^{\frac{1}{3} \cdot 3} = {(x^{\frac{2}{3}})}^{3}$

Step 2.1.2.2

Rewrite the expression.

${(8 x - 7)}^{1} = {(x^{\frac{2}{3}})}^{3}$

Step 2.2

Simplify.

$8 x - 7 = {(x^{\frac{2}{3}})}^{3}$

Step 3

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

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Step 3.1

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

$8 x - 7 = x^{\frac{2}{3} \cdot 3}$

Step 3.2

Cancel the common factor of $3$ .

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Step 3.2.1

Cancel the common factor.

$8 x - 7 = x^{\frac{2}{3} \cdot 3}$

Step 3.2.2

Rewrite the expression.

$8 x - 7 = x^{2}$

Step 4

Subtract $x^{2}$ from both sides of the equation.

$8 x - 7 - x^{2} = 0$

Step 5

Factor the left side of the equation.

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Step 5.1

Factor $- 1$ out of $8 x - 7 - x^{2}$ .

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Step 5.1.1

Reorder the expression.

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Step 5.1.1.1

Move $- 7$ .

$8 x - x^{2} - 7 = 0$

Step 5.1.1.2

Reorder $8 x$ and $- x^{2}$ .

$- x^{2} + 8 x - 7 = 0$

Step 5.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 8 x - 7 = 0$

Step 5.1.3

Factor $- 1$ out of $8 x$ .

$- (x^{2}) - (- 8 x) - 7 = 0$

Step 5.1.4

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

$- (x^{2}) - (- 8 x) - 1 \cdot 7 = 0$

Step 5.1.5

Factor $- 1$ out of $- (x^{2}) - (- 8 x)$ .

$- (x^{2} - 8 x) - 1 \cdot 7 = 0$

Step 5.1.6

Factor $- 1$ out of $- (x^{2} - 8 x) - 1 (7)$ .

$- (x^{2} - 8 x + 7) = 0$

Step 5.2

Factor.

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Step 5.2.1

Factor $x^{2} - 8 x + 7$ using the AC method.

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Step 5.2.1.1

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 $7$ and whose sum is $- 8$ .

$- 7, - 1$

Step 5.2.1.2

Write the factored form using these integers.

$- ((x - 7) (x - 1)) = 0$

Step 5.2.2

Remove unnecessary parentheses.

$- (x - 7) (x - 1) = 0$

Step 6

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

$x - 7 = 0$

$x - 1 = 0$

Step 7

Set $x - 7$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 7$ equal to $0$ .

$x - 7 = 0$

Step 7.2

Add $7$ to both sides of the equation.

$x = 7$

Step 8

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

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Step 8.1

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

$x - 1 = 0$

Step 8.2

Add $1$ to both sides of the equation.

$x = 1$

Step 9

The final solution is all the values that make $- (x - 7) (x - 1) = 0$ true.

$x = 7, 1$