Find the Inverse 10x^(1/3)-6

$10 x^{\frac{1}{3}} - 6$

Interchange the variables.

$x = 10 y^{\frac{1}{3}} - 6$

Solve for $y$ .

Rewrite the equation as $10 y^{\frac{1}{3}} - 6 = x$ .

$10 y^{\frac{1}{3}} - 6 = x$

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

Add $6$ to both sides of the equation.

$10 y^{\frac{1}{3}} - x = 6$

Add $x$ to both sides of the equation.

$10 y^{\frac{1}{3}} = 6 + x$

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

${(10 y^{\frac{1}{3}})}^{3} = {(6 + x)}^{3}$

Simplify the exponent.

Simplify the left side.

Simplify ${(10 y^{\frac{1}{3}})}^{3}$ .

Apply the product rule to $10 y^{\frac{1}{3}}$ .

$10^{3} {(y^{\frac{1}{3}})}^{3} = {(6 + x)}^{3}$

Raise $10$ to the power of $3$ .

$1000 {(y^{\frac{1}{3}})}^{3} = {(6 + x)}^{3}$

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

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

$1000 y^{\frac{1}{3} \cdot 3} = {(6 + x)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

$1000 y^{\frac{1}{3} \cdot 3} = {(6 + x)}^{3}$

Rewrite the expression.

$1000 y^{1} = {(6 + x)}^{3}$

Simplify.

$1000 y = {(6 + x)}^{3}$

Simplify the right side.

Simplify ${(6 + x)}^{3}$ .

Use the Binomial Theorem.

$1000 y = 6^{3} + 3 \cdot 6^{2} x + 3 \cdot 6 x^{2} + x^{3}$

Simplify each term.

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

$1000 y = 216 + 3 \cdot 6^{2} x + 3 \cdot 6 x^{2} + x^{3}$

Raise $6$ to the power of $2$ .

$1000 y = 216 + 3 \cdot 36 x + 3 \cdot 6 x^{2} + x^{3}$

Multiply $3$ by $36$ .

$1000 y = 216 + 108 x + 3 \cdot 6 x^{2} + x^{3}$

Multiply $3$ by $6$ .

$1000 y = 216 + 108 x + 18 x^{2} + x^{3}$

Divide each term in $1000 y = 216 + 108 x + 18 x^{2} + x^{3}$ by $1000$ and simplify.

Divide each term in $1000 y = 216 + 108 x + 18 x^{2} + x^{3}$ by $1000$ .

$\frac{1000 y}{1000} = \frac{216}{1000} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Simplify the left side.

Cancel the common factor of $1000$ .

Cancel the common factor.

$\frac{1000 y}{1000} = \frac{216}{1000} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Divide $y$ by $1$ .

$y = \frac{216}{1000} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Simplify the right side.

Simplify each term.

Cancel the common factor of $216$ and $1000$ .

Factor $8$ out of $216$ .

$y = \frac{8 (27)}{1000} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factors.

Factor $8$ out of $1000$ .

$y = \frac{8 \cdot 27}{8 \cdot 125} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factor.

$y = \frac{8 \cdot 27}{8 \cdot 125} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Rewrite the expression.

$y = \frac{27}{125} + \frac{108 x}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factor of $108$ and $1000$ .

Factor $4$ out of $108 x$ .

$y = \frac{27}{125} + \frac{4 (27 x)}{1000} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factors.

Factor $4$ out of $1000$ .

$y = \frac{27}{125} + \frac{4 (27 x)}{4 (250)} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factor.

$y = \frac{27}{125} + \frac{4 (27 x)}{4 \cdot 250} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Rewrite the expression.

$y = \frac{27}{125} + \frac{27 x}{250} + \frac{18 x^{2}}{1000} + \frac{x^{3}}{1000}$

Cancel the common factor of $18$ and $1000$ .

Factor $2$ out of $18 x^{2}$ .

$y = \frac{27}{125} + \frac{27 x}{250} + \frac{2 (9 x^{2})}{1000} + \frac{x^{3}}{1000}$

Cancel the common factors.

Factor $2$ out of $1000$ .

$y = \frac{27}{125} + \frac{27 x}{250} + \frac{2 (9 x^{2})}{2 (500)} + \frac{x^{3}}{1000}$

Cancel the common factor.

$y = \frac{27}{125} + \frac{27 x}{250} + \frac{2 (9 x^{2})}{2 \cdot 500} + \frac{x^{3}}{1000}$

Rewrite the expression.

$y = \frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}$

Replace $y$ with $f^{- 1} (x)$ to show the final answer.

$f^{- 1} (x) = \frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}$

Verify if $f^{- 1} (x) = \frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}$ is the inverse of $f (x) = 10 x^{\frac{1}{3}} - 6$ .

To verify the inverse, check if $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ .

Evaluate $f^{- 1} (f (x))$ .

Set up the composite result function.

$f^{- 1} (f (x))$

Evaluate $f^{- 1} (10 x^{\frac{1}{3}} - 6)$ by substituting in the value of $f$ into $f^{- 1}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (10 x^{\frac{1}{3}} - 6)}{250} + \frac{9 {(10 x^{\frac{1}{3}} - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Simplify each term.

Factor $2$ out of $27 (10 x^{\frac{1}{3}} - 6)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{2 (27 (5 x^{\frac{1}{3}} - 3))}{250} + \frac{9 {(10 x^{\frac{1}{3}} - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Cancel the common factors.

Factor $2$ out of $250$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{2 (27 (5 x^{\frac{1}{3}} - 3))}{2 (125)} + \frac{9 {(10 x^{\frac{1}{3}} - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Cancel the common factor.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{2 (27 (5 x^{\frac{1}{3}} - 3))}{2 \cdot 125} + \frac{9 {(10 x^{\frac{1}{3}} - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Rewrite the expression.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(10 x^{\frac{1}{3}} - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Simplify the numerator.

Factor $2$ out of $10 x^{\frac{1}{3}} - 6$ .

Factor $2$ out of $10 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(2 (5 x^{\frac{1}{3}}) - 6)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Factor $2$ out of $- 6$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(2 (5 x^{\frac{1}{3}}) + 2 (- 3))}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Factor $2$ out of $2 (5 x^{\frac{1}{3}}) + 2 (- 3)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(2 (5 x^{\frac{1}{3}} - 3))}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Apply the product rule to $2 (5 x^{\frac{1}{3}} - 3)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 \cdot (2^{2} {(5 x^{\frac{1}{3}} - 3)}^{2})}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 \cdot (4 {(5 x^{\frac{1}{3}} - 3)}^{2})}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Multiply $9$ by $4$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{36 {(5 x^{\frac{1}{3}} - 3)}^{2}}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Factor $4$ out of $36 {(5 x^{\frac{1}{3}} - 3)}^{2}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{4 (9 {(5 x^{\frac{1}{3}} - 3)}^{2})}{500} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Cancel the common factors.

Factor $4$ out of $500$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{4 (9 {(5 x^{\frac{1}{3}} - 3)}^{2})}{4 (125)} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Cancel the common factor.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{4 (9 {(5 x^{\frac{1}{3}} - 3)}^{2})}{4 \cdot 125} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Rewrite the expression.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{{(10 x^{\frac{1}{3}} - 6)}^{3}}{1000}$

Simplify the numerator.

Factor $2$ out of $10 x^{\frac{1}{3}} - 6$ .

Factor $2$ out of $10 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{{(2 (5 x^{\frac{1}{3}}) - 6)}^{3}}{1000}$

Factor $2$ out of $- 6$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{{(2 (5 x^{\frac{1}{3}}) + 2 (- 3))}^{3}}{1000}$

Factor $2$ out of $2 (5 x^{\frac{1}{3}}) + 2 (- 3)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{{(2 (5 x^{\frac{1}{3}} - 3))}^{3}}{1000}$

Apply the product rule to $2 (5 x^{\frac{1}{3}} - 3)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{2^{3} {(5 x^{\frac{1}{3}} - 3)}^{3}}{1000}$

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{8 {(5 x^{\frac{1}{3}} - 3)}^{3}}{1000}$

Factor $8$ out of $8 {(5 x^{\frac{1}{3}} - 3)}^{3}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{8 ({(5 x^{\frac{1}{3}} - 3)}^{3})}{1000}$

Cancel the common factors.

Factor $8$ out of $1000$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{8 {(5 x^{\frac{1}{3}} - 3)}^{3}}{8 \cdot 125}$

Cancel the common factor.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{8 {(5 x^{\frac{1}{3}} - 3)}^{3}}{8 \cdot 125}$

Rewrite the expression.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27}{125} + \frac{27 (5 x^{\frac{1}{3}} - 3)}{125} + \frac{9 {(5 x^{\frac{1}{3}} - 3)}^{2}}{125} + \frac{{(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Combine the numerators over the common denominator.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 27 (5 x^{\frac{1}{3}} - 3) + 9 {(5 x^{\frac{1}{3}} - 3)}^{2} + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Simplify each term.

Apply the distributive property.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 27 (5 x^{\frac{1}{3}}) + 27 \cdot - 3 + 9 {(5 x^{\frac{1}{3}} - 3)}^{2} + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $5$ by $27$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} + 27 \cdot - 3 + 9 {(5 x^{\frac{1}{3}} - 3)}^{2} + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $27$ by $- 3$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 {(5 x^{\frac{1}{3}} - 3)}^{2} + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Rewrite ${(5 x^{\frac{1}{3}} - 3)}^{2}$ as $(5 x^{\frac{1}{3}} - 3) (5 x^{\frac{1}{3}} - 3)$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 ((5 x^{\frac{1}{3}} - 3) (5 x^{\frac{1}{3}} - 3)) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Expand $(5 x^{\frac{1}{3}} - 3) (5 x^{\frac{1}{3}} - 3)$ using the FOIL Method.

Apply the distributive property.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 x^{\frac{1}{3}} (5 x^{\frac{1}{3}} - 3) - 3 (5 x^{\frac{1}{3}} - 3)) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Apply the distributive property.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 x^{\frac{1}{3}} (5 x^{\frac{1}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}} - 3)) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Apply the distributive property.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 x^{\frac{1}{3}} (5 x^{\frac{1}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Simplify and combine like terms.

Simplify each term.

Rewrite using the commutative property of multiplication.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 \cdot (5 x^{\frac{1}{3}} x^{\frac{1}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $x^{\frac{1}{3}}$ by $x^{\frac{1}{3}}$ by adding the exponents.

Move $x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 \cdot (5 (x^{\frac{1}{3}} x^{\frac{1}{3}})) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 \cdot (5 x^{\frac{1}{3} + \frac{1}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Combine the numerators over the common denominator.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 \cdot (5 x^{\frac{1 + 1}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Add $1$ and $1$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (5 \cdot (5 x^{\frac{2}{3}}) + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $5$ by $5$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}} + 5 x^{\frac{1}{3}} \cdot - 3 - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $- 3$ by $5$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}} - 15 x^{\frac{1}{3}} - 3 (5 x^{\frac{1}{3}}) - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $5$ by $- 3$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}} - 15 x^{\frac{1}{3}} - 15 x^{\frac{1}{3}} - 3 \cdot - 3) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $- 3$ by $- 3$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}} - 15 x^{\frac{1}{3}} - 15 x^{\frac{1}{3}} + 9) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Subtract $15 x^{\frac{1}{3}}$ from $- 15 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}} - 30 x^{\frac{1}{3}} + 9) + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Apply the distributive property.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 9 (25 x^{\frac{2}{3}}) + 9 (- 30 x^{\frac{1}{3}}) + 9 \cdot 9 + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Simplify.

Multiply $25$ by $9$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} + 9 (- 30 x^{\frac{1}{3}}) + 9 \cdot 9 + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $- 30$ by $9$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 9 \cdot 9 + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Multiply $9$ by $9$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + {(5 x^{\frac{1}{3}} - 3)}^{3}}{125}$

Use the Binomial Theorem.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + {(5 x^{\frac{1}{3}})}^{3} + 3 {(5 x^{\frac{1}{3}})}^{2} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Simplify each term.

Apply the product rule to $5 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 5^{3} {(x^{\frac{1}{3}})}^{3} + 3 {(5 x^{\frac{1}{3}})}^{2} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 {(x^{\frac{1}{3}})}^{3} + 3 {(5 x^{\frac{1}{3}})}^{2} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

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

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x^{\frac{1}{3} \cdot 3} + 3 {(5 x^{\frac{1}{3}})}^{2} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Cancel the common factor of $3$ .

Cancel the common factor.

Rewrite the expression.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 3 {(5 x^{\frac{1}{3}})}^{2} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Simplify.

Apply the product rule to $5 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 3 (5^{2} {(x^{\frac{1}{3}})}^{2}) \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Raise $5$ to the power of $2$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 3 (25 {(x^{\frac{1}{3}})}^{2}) \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Multiply the exponents in ${(x^{\frac{1}{3}})}^{2}$ .

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 3 (25 x^{\frac{1}{3} \cdot 2}) \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Combine $\frac{1}{3}$ and $2$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 3 (25 x^{\frac{2}{3}}) \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Multiply $25$ by $3$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x + 75 x^{\frac{2}{3}} \cdot - 3 + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Multiply $- 3$ by $75$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 3 (5 x^{\frac{1}{3}}) {(- 3)}^{2} + {(- 3)}^{3}}{125}$

Multiply $5$ by $3$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 15 x^{\frac{1}{3}} {(- 3)}^{2} + {(- 3)}^{3}}{125}$

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 15 x^{\frac{1}{3}} \cdot 9 + {(- 3)}^{3}}{125}$

Multiply $9$ by $15$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 135 x^{\frac{1}{3}} + {(- 3)}^{3}}{125}$

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

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 135 x^{\frac{1}{3}} - 27}{125}$

Simplify terms.

Combine the opposite terms in $27 + 135 x^{\frac{1}{3}} - 81 + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 81 + 125 x - 225 x^{\frac{2}{3}} + 135 x^{\frac{1}{3}} - 27$ .

Add $- 81$ and $81$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 0 + 125 x - 225 x^{\frac{2}{3}} + 135 x^{\frac{1}{3}} - 27}{125}$

Add $27 + 135 x^{\frac{1}{3}} + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} + 225 x^{\frac{2}{3}} - 270 x^{\frac{1}{3}} + 125 x - 225 x^{\frac{2}{3}} + 135 x^{\frac{1}{3}} - 27}{125}$

Subtract $225 x^{\frac{2}{3}}$ from $225 x^{\frac{2}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x + 0 + 135 x^{\frac{1}{3}} - 27}{125}$

Add $27 + 135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x$ and $0$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{27 + 135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}} - 27}{125}$

Subtract $27$ from $27$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}} + 0}{125}$

Add $135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}}$ and $0$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{135 x^{\frac{1}{3}} - 270 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}}}{125}$

Subtract $270 x^{\frac{1}{3}}$ from $135 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{- 135 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}}}{125}$

Combine the opposite terms in $- 135 x^{\frac{1}{3}} + 125 x + 135 x^{\frac{1}{3}}$ .

Add $- 135 x^{\frac{1}{3}}$ and $135 x^{\frac{1}{3}}$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{125 x + 0}{125}$

Add $125 x$ and $0$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{125 x}{125}$

Cancel the common factor of $125$ .

Cancel the common factor.

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = \frac{125 x}{125}$

Divide $x$ by $1$ .

$f^{- 1} (10 x^{\frac{1}{3}} - 6) = x$

Evaluate $f (f^{- 1} (x))$ .

Set up the composite result function.

$f (f^{- 1} (x))$

Evaluate $f (\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000})$ by substituting in the value of $f^{- 1}$ into $f$ .

$f (\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}) = 10 {(\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000})}^{\frac{1}{3}} - 6$

Move $\frac{27}{125}$ .

$f (\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}) = 10 {(\frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000} + \frac{27}{125})}^{\frac{1}{3}} - 6$

Move $\frac{27 x}{250}$ .

$f (\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}) = 10 {(\frac{9 x^{2}}{500} + \frac{x^{3}}{1000} + \frac{27 x}{250} + \frac{27}{125})}^{\frac{1}{3}} - 6$

Reorder $\frac{9 x^{2}}{500}$ and $\frac{x^{3}}{1000}$ .

$f (\frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}) = 10 {(\frac{x^{3}}{1000} + \frac{9 x^{2}}{500} + \frac{27 x}{250} + \frac{27}{125})}^{\frac{1}{3}} - 6$

Since $f^{- 1} (f (x)) = x$ and $f (f^{- 1} (x)) = x$ , then $f^{- 1} (x) = \frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}$ is the inverse of $f (x) = 10 x^{\frac{1}{3}} - 6$ .

$f^{- 1} (x) = \frac{27}{125} + \frac{27 x}{250} + \frac{9 x^{2}}{500} + \frac{x^{3}}{1000}$

Find the Inverse 10x^(1/3)-6

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