Question

prove it please I request you ​

Answer 1

Step-by-step explanation:

The formula for (x-y-z)^2(x−y−z)2 is equal to x^2+y^2+z^2-2xy+2yz-2zxx2+y2+z2−2xy+2yz−2zx  

Solution:

Formula for (x-y-z)^2(x−y−z)2

(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+2.x.(−y)+2.(−y).(−z)+2.(−z).x

=x^2+y^2+z^2-2xy+2yz-2zx=x2+y2+z2−2xy+2yz−2zx  

Proof:

Let us take (x-y)=p(x−y)=p

Substituting this in the above given formula,

\begin{gathered}(x-y-z)^2=(p-z)^2\\\\=p^2+z^2-2pz\end{gathered}(x−y−z)2=(p−z)2=p2+z2−2pz

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

\begin{gathered}\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}\end{gathered}=(x−y)2+z2−2pz=x2+y2−2xy+z2−2pz=x2+y2−2xy+z2−2(x−y)z=x2+y2+z2−2xy−2xz+2yz(x−y−z)2=x2+y2+z2−2xy+2yz−2xz

Hence, proved the above driven formula.

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

x^2+y^2+z^2-2xy+2yz-2xzx2+y2+z2−2xy+2yz−2xz