Question

Formula for (x-y-z)2.

Answer 1

Answer:

The formula for $(x-y-z)^2$ is equal to $x^2+y^2+z^2-2xy+2yz-2zx$  

Solution:

Formula for $(x-y-z)^2$

$(x-y-z)^2$  

$=(x)^2+(-y)^2+(z)^2+2.x.(-y)+2.(-y).(-z)+2.(-z).x$

$=x^2+y^2+z^2-2xy+2yz-2zx$  

Proof:

Let us take $(x-y)=p$

Substituting this in the above given formula,

$(x-y-z)^2=(p-z)^2\\\\=p^2+z^2-2pz$

We know that, $(x-y)=p$, substitute this in the above one, we get,

$\begin{array} { c } { = ( x - y ) ^ { 2 } + z ^ { 2 } - 2 p z } \\\\ { = x ^ { 2 } + y ^ { 2 } - 2 x y + z ^ { 2 } - 2 p z } \\\\ { = x ^ { 2 } + y ^ { 2 } - 2 x y + z ^ { 2 } - 2 ( x - y ) z } \\\\ { = x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 x y - 2 x z + 2 y z } \\\\ { ( x - y - z ) ^ { 2 } = x ^ { 2 } + y ^ { 2 } + z ^ { 2 } - 2 x y + 2 y z - 2 x z } \end{array}$

Hence, proved the above driven formula.

Thus, the formula for $(x-y-z)^2$ will be  

$x^2+y^2+z^2-2xy+2yz-2xz$

Answer 2

Answer:

(x-y-z)² = x²+y²+z²-2xy+2yz-2zx

Step-by-step explanation:

(x-y-z)²

= [x+(-y)+(-z)]²

= x²+(-y)²+(-z)²+2x(-y)+2(-y)(-z)+2(-z)x

/* By algebraic identity:

(a+b+c)²=a²+b²+c²+2ab+2bc+2ca */

= x²+y²+z²-2xy+2yz-2zx

Therefore,

(x-y-z)² = x²+y²+z²-2xy+2yz-2zx

•••♪