X^2-Y^2-Z^2: Exploring the Equation

[JDude13]

So
$$x^2-y^2=(x+y)(x-y)$$
in the same sense what does
$$x^2-y^2-z^2=?$$
come to?

 

[chiro]

So
$$x^2-y^2=(x+y)(x-y)$$
in the same sense what does
$$x^2-y^2-z^2=?$$
come to?

You are not going to necessarily get the kind of factorization you got with two terms. If your x^2 - y^2 was itself a positive number, you could apply the same formula that you used for x and y.

If however your x^2 - y^2 was negative you would get a negative term - a negative term which would be in the form -(a + b^2) (a, b^2 >= 0) which has no standard factorization.

 

[JDude13]

Ive had a muck around with it and
$$x^2-y^2-z^2=(x+y+z)(x-y-z)+yz$$

 

[Studiot]

shouldn't that be +2yz?