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) = x^{7} - 4

f (x) = \frac{2 x - 3}{3 x + 4}

y = \frac{7 x ^{2} + 1}{x ^{2} + 7}

y = x e^{x}

f (x) = \frac{3 x - 4}{7 - x}

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