Question

factorise- 9a2 + 4b2 + 16c2 - 6ab - 6bc - 12ac

Answer 1

Step-by-step explanation:

9a2+2+4b2+16c2+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=x3+y3+z3+2xy+2yz+2xz

Therefore, (3a)2+(2b)2+(−4c)2+2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)

= (3a+2b−4c)2