What is the inverse of (x^3+1)/(x^3+2) ?

The inverse of (x^3+1)/(x^3+2) is cube root of (1-2x)/(x-1)

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inverse of (x^3+1)/(x^3+2)

inverse

\frac{x ^{3} + 1}{x ^{3} + 2}

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

Solution steps

y = \frac{x ^{3} + 1}{x ^{3} + 2}

Replace

x

with

y

x = \frac{y ^{3} + 1}{y ^{3} + 2}

Solve for

y, x = \frac{y ^{3} + 1}{y ^{3} + 2}

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

\frac{18}{s ^{4}}

\frac{6 - e ^{- x}}{6 + e ^{- x}}

4 a^{2} - 3 a + 1

4 - 5 x

(2 x + 1)^{\frac{1}{3}}

What is the inverse of (x^3+1)/(x^3+2) ?
The inverse of (x^3+1)/(x^3+2) is cube root of (1-2x)/(x-1)