Solve y=((1-(1-x)^3)/6)^2

$$y = ( ( 1 - ( 1 - x ) ^ { 3 } ) / 6 ) ^ { 2 }$$

Solve for y

$$y=\frac{\left(x\left(x^{2}-3x+3\right)\right)^{2}}{36}$$

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}.

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

To raise \frac{1-\left(1-3x+3x^{2}-x^{3}\right)}{6} to a power, raise both numerator and denominator to the power and then divide.

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

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

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

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

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

Subtract 2 from 1 to get -1.

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

Square 1-3x+3x^{2}-x^{3}.

$$y=\frac{-1+6x-6x^{2}+2x^{3}+x^{6}-6x^{5}+15x^{4}-20x^{3}+15x^{2}-6x+1}{6^{2}}$$

Combine 2x^{3} and -20x^{3} to get -18x^{3}.

$$y=\frac{-1+6x-6x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4}+15x^{2}-6x+1}{6^{2}}$$

Combine -6x^{2} and 15x^{2} to get 9x^{2}.

$$y=\frac{-1+6x+9x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4}-6x+1}{6^{2}}$$

Combine 6x and -6x to get 0.

$$y=\frac{-1+9x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4}+1}{6^{2}}$$

Add -1 and 1 to get 0.

$$y=\frac{9x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4}}{6^{2}}$$

Calculate 6 to the power of 2 and get 36.

$$y=\frac{9x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4}}{36}$$

Divide each term of 9x^{2}-18x^{3}+x^{6}-6x^{5}+15x^{4} by 36 to get \frac{1}{4}x^{2}-\frac{1}{2}x^{3}+\frac{1}{36}x^{6}-\frac{1}{6}x^{5}+\frac{5}{12}x^{4}.

$$y=\frac{1}{4}x^{2}-\frac{1}{2}x^{3}+\frac{1}{36}x^{6}-\frac{1}{6}x^{5}+\frac{5}{12}x^{4}$$

Solve for x

$$\left\{\begin{matrix}x=1\text{, }&y=\frac{1}{36}\\x=-\sqrt[3]{6\sqrt{y}+1}+1\text{; }x=\frac{2^{\frac{2}{3}}\left(\sqrt[3]{2|\frac{6\sqrt{y}-1}{2}|+6\sqrt{y}-1}+\sqrt[3]{-2|\frac{6\sqrt{y}-1}{2}|+6\sqrt{y}-1}+\sqrt[3]{2}\right)}{2}\text{, }&y\geq 0\end{matrix}\right.$$