Question
Factorise:
$9 a^2+3 a-8 b-64 b^2$
 

Answer

$9 a^2+3 a-8 b-64 b^2$
$=9 a^2-64 b^2+3 a-8 b$
$=(3 a)^2-(8 b)^2+(3 a-8 b)\left[\therefore a^2-b^2=(a-b)(a+b)\right]$
$=(3 a+8 b)(3 a-8 b)+(3 a-8 b)$
$=(3 a-8 b)(3 a+8 b+1)$

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