Question

(2x-3y+4z) whole square with identities

Answer 1

By using the identity,

(a+b+c)² = a²+b²+c²+2ab+2bc+2ca

where a=2x, b=-3y, c=4z

we get,

(2x-3y+4z)² = (2x)² + (-3y)² + (4z)² + 2(2x)(-3y) +2(-3y)(4z) + 2(4z)(2x)

= 4x² + 9y² + 16z² - 12xy - 24yz + 16xz

Therefore, (2x-3y+4z)² =4x² + 9y² + 16z² - 12xy - 24yz + 16xz

Hope this is helpful