Factorise: (i) 4x² + 9y² + 16z² + 12xy - 4yz - 16xz (ii) 2x² + yz + 8z² - √2xy + 4√2yz - 8xz

We know that, the algebraic identity: (a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)²
(i) 4x² + 9y² + 16z² + 12xy - 4yz - 16xz
Here, yz and xz terms are negative. i.e., common variable z is in negative form.
= (2x)² + (3y)² + (4z)² + 2(2x)(3y) + 2(3y)(4z) + 2(4z)(2x)
= (2x + 3y - 4z)² [∵ (a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)²]
= (2x + 3y - 4z)(2x + 3y - 4z)
(ii) 2x² + y² + 8z² - √2xy + 4√2yz - 8xz
Here, xy and xz terms are negative. i.e., common variable x is in negative form.
= (-√2x)² + (y)² + (2√2z)² + 2(-√2x)(y) + 2(y)(2√2z) + 2(2√2z)(-√2x)
= (-√2x + y + 2√2z)² [∵ (a² + b² + c² + 2ab + 2bc + 2ca) = (a + b + c)²]
= (-√2x + y + 2√2z)(-√2x + y + 2√2z)