(i) 4x2+9y2+16z2+12xy−24yz−16xz
Using identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca
⇒(2x)2+(3y)2+(−4z)2+2(2x)(3y)+2(3y)(−4z)+2(−4z)(2x)=(2x+3y−4z)2=(2x+3y−4z)(2x+3y−4z)
(ii) 2x2+y2+8z2−2√2xy+4√2yz−8xzUsing the identity (a+b+c)2=a2+b2+c2+2ab+2bc+2ca=(√2x)2+(−y)2+(−2√2z)2+2(√2x)(−y)+2(−y)(−2√2z)+2(√2x)(−2√2z)=(√2x−y−2√2z)2=(√2x−y−2√2z)(√2x−y−2√2z)