What is the inverse of f(x)=(x^3+1)^5 ?

The inverse of f(x)=(x^3+1)^5 is cube root of \sqrt[5]{x}-1

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inverse of f(x)=(x^3+1)^5

inverse

f (x) = (x^{3} + 1)^{5}

Solution steps

y = (x^{3} + 1)^{5}

Replace

x

with

y

x = (y^{3} + 1)^{5}

Solve for

y, x = (y^{3} + 1)^{5}

f (x) = (\frac{1}{7})^{x}

f (x) = \frac{9 x - 17}{8}

f (x) = \frac{3 x + 2}{x - 2}

\frac{x}{1 + x}

\frac{1}{2 x + 1}

What is the inverse of f(x)=(x^3+1)^5 ?
The inverse of f(x)=(x^3+1)^5 is cube root of \sqrt[5]{x}-1