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 $$x,y\text{ and }z$$ .