Solve for x x^(-2/3)+x^(-1/3)-30=0

$x^{- \frac{2}{3}} + x^{- \frac{1}{3}} - 30 = 0$

Factor the left side of the equation.

Rewrite $x^{- \frac{2}{3}}$ as ${(x^{- \frac{1}{3}})}^{2}$ .

${(x^{- \frac{1}{3}})}^{2} + x^{- \frac{1}{3}} - 30 = 0$

Let $u = x^{- \frac{1}{3}}$ . Substitute $u$ for all occurrences of $x^{- \frac{1}{3}}$ .

$u^{2} + u - 30 = 0$

Factor $u^{2} + u - 30$ 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 $- 30$ and whose sum is $1$ .

$- 5, 6$

Write the factored form using these integers.

$(u - 5) (u + 6) = 0$

Replace all occurrences of $u$ with $x^{- \frac{1}{3}}$ .

$(x^{- \frac{1}{3}} - 5) (x^{- \frac{1}{3}} + 6) = 0$

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

$x^{- \frac{1}{3}} - 5 = 0$

$x^{- \frac{1}{3}} + 6 = 0$

Set $x^{- \frac{1}{3}} - 5$ equal to $0$ and solve for $x$ .

Set $x^{- \frac{1}{3}} - 5$ equal to $0$ .

$x^{- \frac{1}{3}} - 5 = 0$

Solve $x^{- \frac{1}{3}} - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$x^{- \frac{1}{3}} = 5$

Raise each side of the equation to the power of $- 3$ to eliminate the fractional exponent on the left side.

${(x^{- \frac{1}{3}})}^{- 3} = 5^{- 3}$

Simplify the exponent.

Simplify the left side.

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

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

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

$x^{- \frac{1}{3} \cdot - 3} = 5^{- 3}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$x^{\frac{- 1}{3} \cdot - 3} = 5^{- 3}$

Factor $3$ out of $- 3$ .

$x^{\frac{- 1}{3} \cdot (3 (- 1))} = 5^{- 3}$

Cancel the common factor.

$x^{\frac{- 1}{3} \cdot (3 \cdot - 1)} = 5^{- 3}$

Rewrite the expression.

$x^{- 1 \cdot - 1} = 5^{- 3}$

Multiply $- 1$ by $- 1$ .

$x^{1} = 5^{- 3}$

Simplify.

$x = 5^{- 3}$

Simplify the right side.

Simplify $5^{- 3}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{5^{3}}$

Raise $5$ to the power of $3$ .

$x = \frac{1}{125}$

Set $x^{- \frac{1}{3}} + 6$ equal to $0$ and solve for $x$ .

Set $x^{- \frac{1}{3}} + 6$ equal to $0$ .

$x^{- \frac{1}{3}} + 6 = 0$

Solve $x^{- \frac{1}{3}} + 6 = 0$ for $x$ .

Subtract $6$ from both sides of the equation.

$x^{- \frac{1}{3}} = - 6$

Raise each side of the equation to the power of $- 3$ to eliminate the fractional exponent on the left side.

${(x^{- \frac{1}{3}})}^{- 3} = {(- 6)}^{- 3}$

Simplify the exponent.

Simplify the left side.

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

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

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

$x^{- \frac{1}{3} \cdot - 3} = {(- 6)}^{- 3}$

Cancel the common factor of $3$ .

Move the leading negative in $- \frac{1}{3}$ into the numerator.

$x^{\frac{- 1}{3} \cdot - 3} = {(- 6)}^{- 3}$

Factor $3$ out of $- 3$ .

$x^{\frac{- 1}{3} \cdot (3 (- 1))} = {(- 6)}^{- 3}$

Cancel the common factor.

$x^{\frac{- 1}{3} \cdot (3 \cdot - 1)} = {(- 6)}^{- 3}$

Rewrite the expression.

$x^{- 1 \cdot - 1} = {(- 6)}^{- 3}$

Multiply $- 1$ by $- 1$ .

$x^{1} = {(- 6)}^{- 3}$

Simplify.

$x = {(- 6)}^{- 3}$

Simplify the right side.

Simplify ${(- 6)}^{- 3}$ .

Rewrite the expression using the negative exponent rule $b^{- n} = \frac{1}{b^{n}}$ .

$x = \frac{1}{{(- 6)}^{3}}$

Raise $- 6$ to the power of $3$ .

$x = \frac{1}{- 216}$

Move the negative in front of the fraction.

$x = - \frac{1}{216}$

The final solution is all the values that make $(x^{- \frac{1}{3}} - 5) (x^{- \frac{1}{3}} + 6) = 0$ true.

$x = \frac{1}{125}, - \frac{1}{216}$

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{125}, - \frac{1}{216}$

Decimal Form:

$x = 0.008, - 0.004\overline{629}$

Solve for x x^(-2/3)+x^(-1/3)-30=0

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