Solve a^3-27/5a^2-16a+3diva^2+3a+a/25a^2-1

$$\frac { a ^ { 3 } - 27 } { 5 a ^ { 2 } - 16 a + 3 } \div \frac { a ^ { 2 } + 3 a + a } { 25 a ^ { 2 } - 1 }$$

Evaluate

$$\frac{\left(5a+1\right)\left(a^{2}+3a+9\right)}{a\left(a+4\right)}$$

Short Solution Steps

Divide \frac{a^{3}-27}{5a^{2}-16a+3} by \frac{a^{2}+3a+a}{25a^{2}-1} by multiplying \frac{a^{3}-27}{5a^{2}-16a+3} by the reciprocal of \frac{a^{2}+3a+a}{25a^{2}-1}.

$$\frac{\left(a^{3}-27\right)\left(25a^{2}-1\right)}{\left(5a^{2}-16a+3\right)\left(a^{2}+3a+a\right)}$$

Combine 3a and a to get 4a.

$$\frac{\left(a^{3}-27\right)\left(25a^{2}-1\right)}{\left(5a^{2}-16a+3\right)\left(a^{2}+4a\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-3\right)\left(5a-1\right)\left(5a+1\right)\left(a^{2}+3a+9\right)}{a\left(a-3\right)\left(5a-1\right)\left(a+4\right)}$$

Cancel out \left(a-3\right)\left(5a-1\right) in both numerator and denominator.

$$\frac{\left(5a+1\right)\left(a^{2}+3a+9\right)}{a\left(a+4\right)}$$

Expand the expression.

$$\frac{5a^{3}+16a^{2}+48a+9}{a^{2}+4a}$$

Expand

$$\frac{5a^{3}+16a^{2}+48a+9}{a^{2}+4a}$$

Short Solution Steps

Divide \frac{a^{3}-27}{5a^{2}-16a+3} by \frac{a^{2}+3a+a}{25a^{2}-1} by multiplying \frac{a^{3}-27}{5a^{2}-16a+3} by the reciprocal of \frac{a^{2}+3a+a}{25a^{2}-1}.

$$\frac{\left(a^{3}-27\right)\left(25a^{2}-1\right)}{\left(5a^{2}-16a+3\right)\left(a^{2}+3a+a\right)}$$

Combine 3a and a to get 4a.

$$\frac{\left(a^{3}-27\right)\left(25a^{2}-1\right)}{\left(5a^{2}-16a+3\right)\left(a^{2}+4a\right)}$$

Factor the expressions that are not already factored.

$$\frac{\left(a-3\right)\left(5a-1\right)\left(5a+1\right)\left(a^{2}+3a+9\right)}{a\left(a-3\right)\left(5a-1\right)\left(a+4\right)}$$

Cancel out \left(a-3\right)\left(5a-1\right) in both numerator and denominator.

$$\frac{\left(5a+1\right)\left(a^{2}+3a+9\right)}{a\left(a+4\right)}$$

Expand the expression.

$$\frac{5a^{3}+16a^{2}+48a+9}{a^{2}+4a}$$