Question

(2x- y + z) whole square

Answer 1

Answer:

Hey friend,

Here is the answer you were looking for:

{(2x - y + z)}^{2} \\ \\ using \: the \: identity \\ {(a - b + c)}^{2} = {a}^{2} + {b}^{2} + {c}^{2} - 2ab - 2bc + 2ca \\ \\ = {(2x)}^{2} + {(y)}^{2} + {(z)}^{2} - 2 \times 2x \times y - 2 \times y \times z + 2 \times z \times 2x \\ \\ = 4 {x}^{2} + {y}^{2} + {z}^{2} - 4xy - 2yz + 4xz

Hope this helps!!!!

Step-by-step explanation:

Answer 2

$\huge\underline\mathfrak\green{AnsweR}$

By using identity V which is

$(a + b + c)^{2} = a ^{2} + b^{2} + c ^{2} + 2ab + bc + 2ca$

$\rule{190}3$

$(2x )^{2} + ( - y)^{2} + (z)^{2} + 2(2x \times - y)2( - y \times z) + 2(z + 2x)$

$= > 4x ^{2} + y^{2} + z^{2} - 4xy - 2yz + 4zx$