What is the identity of (-x+y+z)²
Answers
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Answer:
The formula for (x-y-z)^2(x−y−z)
2
is equal to x^2+y^2+z^2-2xy+2yz-2zxx
2
+y
2
+z
2
−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=x
2
+y
2
+z
2
−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
=p
2
+z
2
−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
+z
2
−2pz
=x
2
+y
2
−2xy+z
2
−2pz
=x
2
+y
2
−2xy+z
2
−2(x−y)z
=x
2
+y
2
+z
2
−2xy−2xz+2yz
(x−y−z)
2
=x
2
+y
2
+z
2
−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-2xzx
2
+y
2
+z
2
−2xy+2yz−2xz