What is the inverse of \sqrt[3]{x^2-x} ?

The inverse of \sqrt[3]{x^2-x} is (1+sqrt(1+4x^3))/2 ,(1-sqrt(1+4x^3))/2

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inverse of \sqrt[3]{x^2-x}

inverse

3 x^{2} - x

\frac{1 + 1 + 4 x ^{3}}{2}, \frac{1 - 1 + 4 x ^{3}}{2}

Solution steps

y = 3 x^{2} - x

Replace

x

with

y

x = 3 y^{2} - y

Solve for

y, x = 3 y^{2} - y

\frac{1 + 1 + 4 x ^{3}}{2}, \frac{1 - 1 + 4 x ^{3}}{2}

f (x) = \frac{- x ^{2}}{16}

f (x) = 3 5 x + 8

f (x) = \frac{4 x + 2}{3 x - 1}

\frac{x}{8} + 3

- 9.56 - 4.95 - 11.95

What is the inverse of \sqrt[3]{x^2-x} ?
The inverse of \sqrt[3]{x^2-x} is (1+sqrt(1+4x^3))/2 ,(1-sqrt(1+4x^3))/2