Question

use identities (-3a+2b-4c)²​

Answer 1

Answer:

We have (x+y+z)

2

=x

2

+y

2

+z

2

+2(xy+yz+zx)

Take x=3a,y=−2b and z=5c

(3a−2b+5c)

2

=(3a)

2

+(−2b)

2

+(5c)

2

+2(3a×−2b+(−2b)×5c+5c×3a)

=9a

2

+4b

2

+25c

2

+2(−6ab−10bc+15ac)

=9a

2

+4b

2

+25c

2

−12ab−20bc+30ac

∴(3a−2b+5c)

2

=9a

2

+4b

2

+25c

2

−12ab−20bc+30ac