Question -
Answer -
(i) 4x2+9y2+16z2+12xy–24yz–16xz
Using identity, (x+y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx= (x+y+z)2
4x2+9y2+16z2+12xy–24yz–16xz =(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–8xz
Using identity, (x +y+z)2 = x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx= (x+y+z)2
2x2+y2+8z2–2√2xy+4√2yz–8xz
= (-√2x)2+(y)2+(2√2z)2+(2×-√2x×y)+(2×y×2√2z)+(2×2√2×−√2x)
= (−√2x+y+2√2z)2
=(−√2x+y+2√2z)(−√2x+y+2√2z)