Question

What is the formula of \[{x^2} + {y^2} + {z^2}?\]

Answer 1

Hint: We use some algebraic identities to solve this problem. We also know the formula \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] and we are going to use this formula to solve this problem. We also use the concepts of transpositions from one side to 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 solve this problem.

Complete step by step solution:
To solve this problem, first we need to know the way to expand the squares.
Let us first evaluate the value of \[{(a + b)^2}\] which is equal to \[(a + b)(a + b)\] .
\[ \Rightarrow (a + b)(a + b) = a.a + a.b + b.a + b.b\]
\[ \Rightarrow (a + b)(a + b) = {a^2} + 2ab + {b^2}\]
So, we can conclude that, \[{(a + b)^2} = {a^2} + {b^2} + 2ab\] -----equation (1)
So, now let us evaluate \[{(x + y + z)^2}\]
Here, in the term \[x + y + z\] , we will group two terms into a single term as \[((x + y) + z)\]
So, \[{(x + y + z)^2} = {((x + y) + z)^2}\]
And this is of the form \[{(a + b)^2}\] 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,
\[{(x + y + z)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2xz + 2yz\]
Now, we use transpositions to get the value of \[{x^2} + {y^2} + {z^2}\] .
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {(x + y + z)^2} - 2xy - 2xz - 2yz\]
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 the other side and vice-versa. And similarly, the multiplication changes to division when transposed to 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,
\[{(x - y - z)^2} = {x^2} + {y^2} + {z^2} - 2xy - 2xz + 2yz\]
\[ \Rightarrow {x^2} + {y^2} + {z^2} = {(x - y - z)^2} + 2xy + 2xz - 2yz\]
So, this can also be the value of it.
The other forms are,
\[{x^2} + {y^2} + {z^2} = {(x + y - z)^2} - 2xy + 2xz + 2yz\]
And
\[{x^2} + {y^2} + {z^2} = {(x - y + z)^2} + 2xy - 2xz + 2yz\]
We can also get many forms by changing the signs of \[x,y{\text{ and }}z\] .