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

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

Solve for y

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

Solution Steps

Anything divided by one gives itself.

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

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

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

Calculate -x to the power of 2 and get x^{2}.

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

To find the opposite of 1+3\left(-x\right)+3x^{2}+\left(-x\right)^{3}, find the opposite of each term.

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

Subtract 1 from 1 to get 0.

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

Multiply -3 and -1 to get 3.

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

Use the distributive property to multiply \frac{1}{6} by 3x-3x^{2}-\left(-x\right)^{3}.

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

Expand \left(-x\right)^{3}.

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

Calculate -1 to the power of 3 and get -1.

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

Multiply -\frac{1}{6} and -1 to get \frac{1}{6}.

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

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

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