Solve for x x^(-2/3)+x^(-1/3)-6=0

$x^{- \frac{2}{3}} + x^{- \frac{1}{3}} - 6 = 0$

Step 1

Factor the left side of the equation.

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

${(x^{- \frac{1}{3}})}^{2} + x^{- \frac{1}{3}} - 6 = 0$

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

$u^{2} + u - 6 = 0$

Factor $u^{2} + u - 6$ 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 $- 6$ and whose sum is $1$ .

$- 2, 3$

Write the factored form using these integers.

$(u - 2) (u + 3) = 0$

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

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

Step 2

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

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

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

Step 3

Set $x^{- \frac{1}{3}} - 2$ equal to $0$ and solve for $x$ .

Set $x^{- \frac{1}{3}} - 2$ equal to $0$ .

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

Solve $x^{- \frac{1}{3}} - 2 = 0$ for $x$ .

Add $2$ to both sides of the equation.

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

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

$x^{- \frac{1}{3} \cdot - 3} = 2^{- 3}$

Cancel the common factor of $3$ .

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

$x^{\frac{- 1}{3} \cdot - 3} = 2^{- 3}$

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

$x^{- 1 \cdot - 1} = 2^{- 3}$

Multiply $- 1$ by $- 1$ .

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

Simplify.

$x = 2^{- 3}$

Simplify the right side.

Simplify $2^{- 3}$ .

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

$x = \frac{1}{2^{3}}$

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

$x = \frac{1}{8}$

Step 4

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

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

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

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

Subtract $3$ from both sides of the equation.

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

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

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

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, ${(a^{m})}^{n} = a^{m n}$ .

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

Cancel the common factor of $3$ .

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

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

Factor $3$ out of $- 3$ .

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

Cancel the common factor.

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

Rewrite the expression.

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

Multiply $- 1$ by $- 1$ .

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

Simplify.

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

Simplify the right side.

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

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

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

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

$x = \frac{1}{- 27}$

Move the negative in front of the fraction.

$x = - \frac{1}{27}$

Step 5

The final solution is all the values that make $(x^{- \frac{1}{3}} - 2) (x^{- \frac{1}{3}} + 3) = 0$ true.

$x = \frac{1}{8}, - \frac{1}{27}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1}{8}, - \frac{1}{27}$

Decimal Form:

$x = 0.125, - 0.\overline{037}$

Solve for x x^(-2/3)+x^(-1/3)-6=0

Related Questions

Support LCMGCF.com by disabling your adblocker.