Question

factorise 9a2+4b2+c2-12ab-4bc+6ca

Answer 1

Answer:

9a

2

+4b

2

+16c

2

+2ab−16bc−24ca

= (3a)

2

+(2b)

2

+(−4c)

2

+2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)

Suitable identities is (x+y+z)

3

=x

3

+y

3

+z

3

+2xy+2yz+2xz

Therefore, (3a)

2

+(2b)

2

+(−4c)

2

+2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)

= (3a+2b−4c)

2