Solve y=1/6*(1-(1-x)^3)

$$y = 1 / 6 \cdot ( 1 - ( 1 - x ) ^ { 3 } )$$

Solve for y

$$y=-\frac{x\left(-x^{2}+3x-3\right)}{6}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{3}=a^{3}-3a^{2}b+3ab^{2}-b^{3} to expand \left(1-x\right)^{3}.

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

To find the opposite of 1-3x+3x^{2}-x^{3}, find the opposite of each term.

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

Subtract 1 from 1 to get 0.

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

Use the distributive property to multiply \frac{1}{6} by 3x-3x^{2}+x^{3}.

$$y=\frac{1}{2}x-\frac{1}{2}x^{2}+\frac{1}{6}x^{3}$$

Solve for x (complex solution)

$$x=-\sqrt[3]{1-6y}+1$$

$$x=e^{\frac{\pi i}{3}}\sqrt[3]{1-6y}+1$$

$$x=e^{\frac{5i\pi }{3}}\sqrt[3]{1-6y}+1$$

Solve for x

$$x=-\sqrt[3]{1-6y}+1$$