Solve 1-y/2+y^3/6=c

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

Solve for c

$$c=\frac{y^{3}-3y+6}{6}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Least common multiple of 2 and 6 is 6. Multiply \frac{y}{2} times \frac{3}{3}.

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

Since -\frac{3y}{6} and \frac{y^{3}}{6} have the same denominator, add them by adding their numerators.

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

Divide each term of -3y+y^{3} by 6 to get -\frac{1}{2}y+\frac{1}{6}y^{3}.

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

Swap sides so that all variable terms are on the left hand side.

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