Multiply:
9x2+25y2+15xy−12x+20y+16
by 3x−5y+4
To multiply: 9x2+25y2+15xy−12x+20y+16
by 3x−5y+4
∴(3x−5y+4)×(9x2+25y2+15xy−12x+20y+16)
=(3x−5y+4)×(9x2+25y2+16+15xy+20y−12x)
=(3x−5y+4)×[(3x)2+(−5y)2+(4)2
−3x×(−5y)−(−5y)×(4)−(3x)×(4)]
=(3x)3+(−5y)3+(4)3−3×3x×(−5y)×(4)
[∵(a+b+c)(a2+b2+c2−ab−bc−ca)=a3+b3+c3−3abc]
=27x3−125y3+64+180xy
Hence, 27x3−125y3+64+180xy is the required result.