What is the inverse of f(x)=(x^3+1)^4 ?

The inverse of f(x)=(x^3+1)^4 is cube root of \sqrt[4]{x}-1},\sqrt[3]{-\sqrt[4]{x}-1

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inverse of f(x)=(x^3+1)^4

inverse

f (x) = (x^{3} + 1)^{4}

Solution steps

y = (x^{3} + 1)^{4}

Replace

x

with

y

x = (y^{3} + 1)^{4}

Solve for

y, x = (y^{3} + 1)^{4}

f (x) = arccoth (x)

f (x) = \frac{2 x}{x - 1}

f (x) = 2 x^{2}, x \geq 0

f (x) = 7 x^{3} - 12

f (x) = \frac{4}{5 x + 3}, x \neq = - \frac{3}{5}

What is the inverse of f(x)=(x^3+1)^4 ?
The inverse of f(x)=(x^3+1)^4 is cube root of \sqrt[4]{x}-1},\sqrt[3]{-\sqrt[4]{x}-1