Factorise using identity 9a²+4b²+c²-12ab-4bc+6ac
Answers
Answer:
9a
9a 2
9a 2 +4b
9a 2 +4b 2
9a 2 +4b 2 +16c
9a 2 +4b 2 +16c 2
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a)
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2 +(2b)
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2 +(2b) 2
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2 +(2b) 2 +(−4c)
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2 +(2b) 2 +(−4c) 2
9a 2 +4b 2 +16c 2 +2ab−16bc−24ca= (3a) 2 +(2b) 2 +(−4c) 2 +2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)
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)
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
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
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
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
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
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
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
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
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+2xzTherefore, (3a)
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+2xzTherefore, (3a) 2
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+2xzTherefore, (3a) 2 +(2b)
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+2xzTherefore, (3a) 2 +(2b) 2
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+2xzTherefore, (3a) 2 +(2b) 2 +(−4c)
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+2xzTherefore, (3a) 2 +(2b) 2 +(−4c) 2
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+2xzTherefore, (3a) 2 +(2b) 2 +(−4c) 2 +2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)
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+2xzTherefore, (3a) 2 +(2b) 2 +(−4c) 2 +2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)= (3a+2b−4c)
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+2xzTherefore, (3a) 2 +(2b) 2 +(−4c) 2 +2(3a)(2b)+2(2b)(−4c)+2(−4c)(2a)= (3a+2b−4c) 2
hope it's helpful
❤️badmash ladka❤️
Answer:
(3a-2b+c)²
Step-by-step explanation:
we know that,
a²+b²+c²+2ab+2bc+2ac =(a+b+c)².
9a²+4b²+c²-12ab-4bc+6ac
=(3a)²+(-2b)²+(1c)²+2(-6)ab+2(-2)bc+2(3)ac
=(3a-2b+1c)²