Solve (a^2+a+1-a^3/a-1)(1+a-4a/1+a)

$$( a ^ { 2 } + a + 1 - \frac { a ^ { 3 } } { a - 1 } ) ( 1 + a - \frac { 4 a } { 1 + a } )$$

Evaluate

$$-\frac{a-1}{a+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2}+a+1 times \frac{a-1}{a-1}.

$$\left(\frac{\left(a^{2}+a+1\right)\left(a-1\right)}{a-1}-\frac{a^{3}}{a-1}\right)\left(1+a-\frac{4a}{1+a}\right)$$

Since \frac{\left(a^{2}+a+1\right)\left(a-1\right)}{a-1} and \frac{a^{3}}{a-1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a^{2}+a+1\right)\left(a-1\right)-a^{3}}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

Do the multiplications in \left(a^{2}+a+1\right)\left(a-1\right)-a^{3}.

$$\frac{a^{3}-a^{2}+a^{2}-a+a-1-a^{3}}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

Combine like terms in a^{3}-a^{2}+a^{2}-a+a-1-a^{3}.

$$\frac{-1}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

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

$$\frac{-1}{a-1}\left(\frac{\left(1+a\right)\left(1+a\right)}{1+a}-\frac{4a}{1+a}\right)$$

Since \frac{\left(1+a\right)\left(1+a\right)}{1+a} and \frac{4a}{1+a} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-1}{a-1}\times \frac{\left(1+a\right)\left(1+a\right)-4a}{1+a}$$

Do the multiplications in \left(1+a\right)\left(1+a\right)-4a.

$$\frac{-1}{a-1}\times \frac{1+a+a+a^{2}-4a}{1+a}$$

Combine like terms in 1+a+a+a^{2}-4a.

$$\frac{-1}{a-1}\times \frac{1-2a+a^{2}}{1+a}$$

Multiply \frac{-1}{a-1} times \frac{1-2a+a^{2}}{1+a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(1-2a+a^{2}\right)}{\left(a-1\right)\left(1+a\right)}$$

Factor the expressions that are not already factored.

$$\frac{-\left(a-1\right)^{2}}{\left(a-1\right)\left(a+1\right)}$$

Cancel out a-1 in both numerator and denominator.

$$\frac{-\left(a-1\right)}{a+1}$$

Expand the expression.

$$\frac{-a+1}{a+1}$$

Expand

$$-\frac{a-1}{a+1}$$

Short Solution Steps

To add or subtract expressions, expand them to make their denominators the same. Multiply a^{2}+a+1 times \frac{a-1}{a-1}.

$$\left(\frac{\left(a^{2}+a+1\right)\left(a-1\right)}{a-1}-\frac{a^{3}}{a-1}\right)\left(1+a-\frac{4a}{1+a}\right)$$

Since \frac{\left(a^{2}+a+1\right)\left(a-1\right)}{a-1} and \frac{a^{3}}{a-1} have the same denominator, subtract them by subtracting their numerators.

$$\frac{\left(a^{2}+a+1\right)\left(a-1\right)-a^{3}}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

Do the multiplications in \left(a^{2}+a+1\right)\left(a-1\right)-a^{3}.

$$\frac{a^{3}-a^{2}+a^{2}-a+a-1-a^{3}}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

Combine like terms in a^{3}-a^{2}+a^{2}-a+a-1-a^{3}.

$$\frac{-1}{a-1}\left(1+a-\frac{4a}{1+a}\right)$$

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

$$\frac{-1}{a-1}\left(\frac{\left(1+a\right)\left(1+a\right)}{1+a}-\frac{4a}{1+a}\right)$$

Since \frac{\left(1+a\right)\left(1+a\right)}{1+a} and \frac{4a}{1+a} have the same denominator, subtract them by subtracting their numerators.

$$\frac{-1}{a-1}\times \frac{\left(1+a\right)\left(1+a\right)-4a}{1+a}$$

Do the multiplications in \left(1+a\right)\left(1+a\right)-4a.

$$\frac{-1}{a-1}\times \frac{1+a+a+a^{2}-4a}{1+a}$$

Combine like terms in 1+a+a+a^{2}-4a.

$$\frac{-1}{a-1}\times \frac{1-2a+a^{2}}{1+a}$$

Multiply \frac{-1}{a-1} times \frac{1-2a+a^{2}}{1+a} by multiplying numerator times numerator and denominator times denominator.

$$\frac{-\left(1-2a+a^{2}\right)}{\left(a-1\right)\left(1+a\right)}$$

Factor the expressions that are not already factored.

$$\frac{-\left(a-1\right)^{2}}{\left(a-1\right)\left(a+1\right)}$$

Cancel out a-1 in both numerator and denominator.

$$\frac{-\left(a-1\right)}{a+1}$$

Expand the expression.

$$\frac{-a+1}{a+1}$$