Solve (3+k)^2/9+3k*6-6k/9-k^2

$$\frac { ( 3 + k ) ^ { 2 } } { 9 + 3 k } \times \frac { 6 - 6 k } { 9 - k ^ { 2 } }$$

Evaluate

$$-\frac{2\left(1-k\right)}{k-3}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{\left(3+k\right)^{2}}{9+3k}.

$$\frac{\left(k+3\right)^{2}}{3\left(k+3\right)}\times \frac{6-6k}{9-k^{2}}$$

Cancel out k+3 in both numerator and denominator.

$$\frac{k+3}{3}\times \frac{6-6k}{9-k^{2}}$$

Multiply \frac{k+3}{3} times \frac{6-6k}{9-k^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(k+3\right)\left(6-6k\right)}{3\left(9-k^{2}\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in 3+k.

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

Cancel out 3\left(-k-3\right) in both numerator and denominator.

$$\frac{-2\left(-k+1\right)}{k-3}$$

Expand the expression.

$$\frac{2k-2}{k-3}$$

Expand

$$-\frac{2\left(k-1\right)}{3-k}$$

Short Solution Steps

Factor the expressions that are not already factored in \frac{\left(3+k\right)^{2}}{9+3k}.

$$\frac{\left(k+3\right)^{2}}{3\left(k+3\right)}\times \frac{6-6k}{9-k^{2}}$$

Cancel out k+3 in both numerator and denominator.

$$\frac{k+3}{3}\times \frac{6-6k}{9-k^{2}}$$

Multiply \frac{k+3}{3} times \frac{6-6k}{9-k^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(k+3\right)\left(6-6k\right)}{3\left(9-k^{2}\right)}$$

Factor the expressions that are not already factored.

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

Extract the negative sign in 3+k.

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

Cancel out 3\left(-k-3\right) in both numerator and denominator.

$$\frac{-2\left(-k+1\right)}{k-3}$$

Expand the expression.

$$\frac{2k-2}{k-3}$$