Question

Simplify:

\[\frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]

Answer 1

It is known that,

a² − b² = (a + b) (a − b)

a³ − b³ = (a − b)(a² + ab + b²)

\[\  \frac{a^3 - 27}{5 a^2 - 16a + 3} \div \frac{a^2 + 3a + 9}{25 a^2 - 1}\]

\[ = \frac{\left( a \right)^3 - \left(3 \right)^3}{5 a^2 - 15a - a + 3}   \div \frac{a^2 + 3a + 9}{\left( 5a \right)^2 - \left( 1 \right)^2}\]

\[ = \frac{\left(a - 3 \right)\left\{ \left( a \right)^2 + \left(a \right) \times \left(3 \right) + \left(3 \right)^2 \right\}}{5a\left(a - 3 \right) - 1\left( a - 3 \right)} \div \frac{a^2 + 3a + 9}{\left(5a + 1 \right)\left(5a - 1 \right)}\]

\[ = \frac{\left(a - 3 \right)\left(a^2 + 3a + 9 \right)}{\left(5a - 1 \right)\left(a - 3 \right)} \div \frac{\left( a^2 + 3a + 9 \right)}{\left(5a + 1 \right)\left(5a - 1 \right)}\]

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

\[ = 5a + 1\]