What is the inverse of f(x)=(x^3-3)^{1/3-1} ?

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

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inverse of f(x)=(x^3-3)^{1/3-1}

inverse

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

Solution steps

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

Replace

x

with

y

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

Solve for

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

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