# x^2+y^2+z^2

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Hint: We use some algebraic identities to tướng solve this problem. We also know the formula ${(a + b)^2} = {a^2} + {b^2} + 2ab$ and we are going to tướng use this formula to tướng solve this problem. We also use the concepts of transpositions from one side to tướng another.
And, ${(a + b)^2} = {a^2} + {b^2} + 2ab$ $\Rightarrow {a^2} + {b^2} = {(a + b)^2} - 2ab$
If we change the signs of $a$ and $b$ , we get a few other forms and we use this concept also to tướng solve this problem.

Complete step by step solution:
To solve this problem, first we need to tướng know the way to tướng expand the squares.
Let us first evaluate the value of ${(a + b)^2}$ which is equal to tướng $(a + b)(a + b)$ .
$\Rightarrow (a + b)(a + b) = a.a + a.b + b.a + b.b$
$\Rightarrow (a + b)(a + b) = {a^2} + 2ab + {b^2}$
So, we can conclude that, ${(a + b)^2} = {a^2} + {b^2} + 2ab$ -----equation (1)
So, now let us evaluate ${(x + hắn + z)^2}$
Here, in the term $x + hắn + z$ , we will group two terms into a single term as $((x + y) + z)$
So, ${(x + hắn + z)^2} = {((x + y) + z)^2}$
And this is of the size ${(a + b)^2}$ and $a = x + y$ and $b = z$ .
So, we can write it as,
${((x + y) + z)^2} = {(x + y)^2} + {z^2} + 2(x + y)z$ ------from equation (1)
$\Rightarrow {((x + y) + z)^2} = {(x + y)^2} + {z^2} + 2xz + 2yz$
Now, let’s simplify further
$\Rightarrow {((x + y) + z)^2} = {x^2} + {y^2} + 2xy + {z^2} + 2xz + 2yz$
So, finally on rearranging terms, we get,
${(x + hắn + z)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2xz + 2yz$
Now, we use transpositions to tướng get the value of ${x^2} + {y^2} + {z^2}$ .
$\Rightarrow {x^2} + {y^2} + {z^2} = {(x + hắn + z)^2} - 2xy - 2xz - 2yz$
So, this is the formula for ${x^2} + {y^2} + {z^2}$ .

Note: Make sure that you perform transpositions correctly. In transpositions, the positive terms become negative when transposed to tướng the other side and vice-versa. And similarly, the multiplication changes to tướng division when transposed to tướng the other side and vice-versa. While grouping terms, we grouped $x$ and $y$ terms. But instead, we can also group the terms $y$ and $z$ and we will still get the same result.
Be careful by expanding the squares.
And also,
${(x - hắn - z)^2} = {x^2} + {y^2} + {z^2} - 2xy - 2xz + 2yz$
$\Rightarrow {x^2} + {y^2} + {z^2} = {(x - hắn - z)^2} + 2xy + 2xz - 2yz$
So, this can also be the value of it.
The other forms are,
${x^2} + {y^2} + {z^2} = {(x + hắn - z)^2} - 2xy + 2xz + 2yz$
And
${x^2} + {y^2} + {z^2} = {(x - hắn + z)^2} + 2xy - 2xz + 2yz$
We can also get many forms by changing the signs of $x,y{\text{ and }}z$ .

Last updated date: 06th Sep 2023

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