Using suitable identity multiply:

9x² + 25y² + 15xy + 12x – 20y + 16 by 3x – 5y – 4

Given: 9x² + 25y² + 15xy + 12x – 20y + 16 and 3x –5y– 4

So, we have,

(3x – 5y – 4) (9x² + 25y² + 15xy + 12x – 20y + 16)

[3x + (– 5y) + (– 4)] [(3x)² + (– 5y)² + (– 4)² – (3x) (– 5y) – (– 5y) (– 4) – (– 4) (3x)]

Using identity:

a³ + b³ + c³ – 3abc = (a + b + c) (a² +b² +c² –ab –bc– ca)

We have,

= (3x)³ + (– 5y)³ + (– 4)³ – 3(3x) (– 5y) (– 4)

= 27x³ – 125y³ – 64 – 180xy