Question

a³-27/5a²-16a+3 ÷ a²+3a+9/25a²-1
Please help me in solving this sum it is very difficult.
standard 8th.​

Answer 1

$\frac{ {a}^{3} - 27 }{5 {a}^{2} - 16a + 3 } \div \frac{ {a}^{2} + 3a + 9}{25 {a}^{2} - 1 } \\ ( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} ) = \\ \frac{ {a}^{3} - 27 }{5 {a}^{2} - 16a + 3 } \times \frac{{25 {a}^{2} - 1 }}{ {a}^{2} + 3a + 9}{} = \\ \frac{(a - 3)( {a}^{2} + 3a + a)}{(5a - 1)(5a + 1)} \times \frac{(5a - 1)(5a + 1)}{ {a}^{2} + 4a + a } \\ = 5a + 1$

Answer 2

Answer:

But I already stop to reporting.

$\frac{ {a}^{3} - 27 }{5 {a}^{2} - 16a + 3 } \div \frac{ {a}^{2} + 3a + 9}{25 {a}^{2} - 1 } \\ ( \frac{a}{b} \div \frac{c}{d} = \frac{a}{b} \times \frac{d}{c} ) = \\ \frac{ {a}^{3} - 27 }{5 {a}^{2} - 16a + 3 } \times \frac{{25 {a}^{2} - 1 }}{ {a}^{2} + 3a + 9}{} = \\ \frac{(a - 3)( {a}^{2} + 3a + a)}{(5a - 1)(5a + 1)} \times \frac{(5a - 1)(5a + 1)}{ {a}^{2} + 4a + a } \\ = 5a + 1$

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