What is the inverse of (x^3-8)/(x-2) ?

The inverse of (x^3-8)/(x-2) is -1+sqrt(x-3),-sqrt(x-3)-1

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inverse of (x^3-8)/(x-2)

inverse

\frac{x ^{3} - 8}{x - 2}

- 1 + x - 3, - x - 3 - 1

Solution steps

y = \frac{x ^{3} - 8}{x - 2}

Replace

x

with

y

x = \frac{y ^{3} - 8}{y - 2}

Solve for

y, x = \frac{y ^{3} - 8}{y - 2}

- 1 + x - 3, - x - 3 - 1

y = 6 - (- \frac{2}{3} cos (x))^{2} + 1

\frac{4 x - 5}{2 x + 1}

\frac{x ^{2} + 2 x + 1}{x ^{2} - 1}

\frac{2 x + 1}{2 x - 1}

y = lo g_{5} (x + 3) - 8

What is the inverse of (x^3-8)/(x-2) ?
The inverse of (x^3-8)/(x-2) is -1+sqrt(x-3),-sqrt(x-3)-1