Question

Simplify:

$$\frac{a^3-27}{5a^2-16a+3}÷\frac{a^2+3a+9}{25a^2-1}$$

Answer 1

It is known that,

a² − b² = (a + b) (a − b)

a³ − b³ = (a − b)(a² + ab + b²)

$$\frac{a^3-27}{5a^2-16a+3}÷\frac{a^2+3a+9}{25a^2-1}$$

$$=\frac{{\left(a\right)}^3-{\left(3\right)}^3}{5a^2-15a-a+3}÷\frac{a^2+3a+9}{{\left(5a\right)}^2-{\left(1\right)}^2}$$

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

$$=\frac{(a-3)(a^2+3a+9)}{(5a-1)(a-3)}÷\frac{(a^2+3a+9)}{(5a+1)(5a-1)}$$

$$=\frac{(a-3)(a^2+3a+9)}{(5a-1)(a-3)}\times \frac{(5a+1)(5a-1)}{(a^2+3a+9)}$$

$$=5a+1$$