Solve a^3+8/4a^2-1*1-2ra/2a^2-4a+8*6a+3/4+4a+a^2

$$\frac { a ^ { 3 } + 8 } { 4 a ^ { 2 } - 1 } \cdot \frac { 1 - 2 r a } { 2 a ^ { 2 } - 4 a + 8 } \cdot \frac { 6 a + 3 } { 4 + 4 a + a ^ { 2 } }$$

Evaluate

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

Short Solution Steps

Multiply \frac{a^{3}+8}{4a^{2}-1} times \frac{1-2ra}{2a^{2}-4a+8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{3}+8\right)\left(1-2ra\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)}\times \frac{6a+3}{4+4a+a^{2}}$$

Multiply \frac{\left(a^{3}+8\right)\left(1-2ra\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)} times \frac{6a+3}{4+4a+a^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{3}+8\right)\left(1-2ra\right)\left(6a+3\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)\left(4+4a+a^{2}\right)}$$

Factor the expressions that are not already factored.

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

Cancel out \left(a+2\right)\left(2a+1\right)\left(a^{2}-2a+4\right) in both numerator and denominator.

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

Expand the expression.

$$\frac{-6ar+3}{4a^{2}+6a-4}$$

Expand

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

Short Solution Steps

Multiply \frac{a^{3}+8}{4a^{2}-1} times \frac{1-2ra}{2a^{2}-4a+8} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{3}+8\right)\left(1-2ra\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)}\times \frac{6a+3}{4+4a+a^{2}}$$

Multiply \frac{\left(a^{3}+8\right)\left(1-2ra\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)} times \frac{6a+3}{4+4a+a^{2}} by multiplying numerator times numerator and denominator times denominator.

$$\frac{\left(a^{3}+8\right)\left(1-2ra\right)\left(6a+3\right)}{\left(4a^{2}-1\right)\left(2a^{2}-4a+8\right)\left(4+4a+a^{2}\right)}$$

Factor the expressions that are not already factored.

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

Cancel out \left(a+2\right)\left(2a+1\right)\left(a^{2}-2a+4\right) in both numerator and denominator.

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

Expand the expression.

$$\frac{-6ar+3}{4a^{2}+6a-4}$$