Solve 1+x-1x^3/3/1+x-frac{x^3{6}}

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

Evaluate

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+x times \frac{3}{3}.

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

Since \frac{3\left(1+x\right)}{3} and \frac{x^{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(1+x\right)-x^{3}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+x times \frac{6}{6}.

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

Since \frac{6\left(1+x\right)}{6} and \frac{x^{3}}{6} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 6\left(1+x\right)-x^{3}.

$$\frac{\frac{3+3x-x^{3}}{3}}{\frac{6+6x-x^{3}}{6}}$$

Divide \frac{3+3x-x^{3}}{3} by \frac{6+6x-x^{3}}{6} by multiplying \frac{3+3x-x^{3}}{3} by the reciprocal of \frac{6+6x-x^{3}}{6}.

$$\frac{\left(3+3x-x^{3}\right)\times 6}{3\left(6+6x-x^{3}\right)}$$

Cancel out 3 in both numerator and denominator.

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

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

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

Expand

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

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+x times \frac{3}{3}.

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

Since \frac{3\left(1+x\right)}{3} and \frac{x^{3}}{3} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 3\left(1+x\right)-x^{3}.

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

To add or subtract expressions, expand them to make their denominators the same. Multiply 1+x times \frac{6}{6}.

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

Since \frac{6\left(1+x\right)}{6} and \frac{x^{3}}{6} have the same denominator, subtract them by subtracting their numerators.

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

Do the multiplications in 6\left(1+x\right)-x^{3}.

$$\frac{\frac{3+3x-x^{3}}{3}}{\frac{6+6x-x^{3}}{6}}$$

Divide \frac{3+3x-x^{3}}{3} by \frac{6+6x-x^{3}}{6} by multiplying \frac{3+3x-x^{3}}{3} by the reciprocal of \frac{6+6x-x^{3}}{6}.

$$\frac{\left(3+3x-x^{3}\right)\times 6}{3\left(6+6x-x^{3}\right)}$$

Cancel out 3 in both numerator and denominator.

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

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

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