Question

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

• A. $5a-1$
• B. $5a+1$
• C. $a+1$
• D. $a-1$

Answer 1

Answer: (B)

$5a+1$

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

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

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

$=(5a+1)$