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

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

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inverse of (x^3+1)^4

inverse

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

3 4 x - 1, 3 - 4 x - 1

Solution steps

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

Replace

x

with

y

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

Solve for

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

3 4 x - 1, 3 - 4 x - 1

\frac{6 x - 7}{9 x + 6}

\frac{x + 3}{x + 6}

f (x) = \frac{( 2 x - 3 )}{x - 1}

f (x) = \frac{7 x + 9}{4 x - 7}

f (x) = - 3 - 8 x + 8 - 10

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