Question -
Answer -
(i) 4x2+9y2+16z2+12xyтАУ24yzтАУ16xz
Using identity, (x+y+z)2┬а= x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx= (x+y+z)2
4x2+9y2+16z2+12xyтАУ24yzтАУ16xz┬а=(2x)2+(3y)2+(тИТ4z)2+(2├Ч2x├Ч3y)+(2├Ч3y├ЧтИТ4z)+(2├ЧтИТ4z├Ч2x)
= (2x+3yтАУ4z)2
= (2x+3yтАУ4z)(2x+3yтАУ4z)
(ii) 2x2+y2+8z2тАУ2тИЪ2xy+4тИЪ2yzтАУ8xz
Using identity, (x +y+z)2┬а= x2+y2+z2+2xy+2yz+2zx
We can say that, x2+y2+z2+2xy+2yz+2zx= (x+y+z)2
2x2+y2+8z2тАУ2тИЪ2xy+4тИЪ2yzтАУ8xz
= (-тИЪ2x)2+(y)2+(2тИЪ2z)2+(2├Ч-тИЪ2x├Чy)+(2├Чy├Ч2тИЪ2z)+(2├Ч2тИЪ2├ЧтИТтИЪ2x)
= (тИТтИЪ2x+y+2тИЪ2z)2
=(тИТтИЪ2x+y+2тИЪ2z)(тИТтИЪ2x+y+2тИЪ2z)