#3**0 **

so no because if, lets say x is 2. y could be -2 and the equation would still work

Guest Jan 29, 2021

#4**0 **

POSSIBLY

rember when yo take the square root of some thing you get a plus and a minus root

sqrt 4 = ±2

so x^{2} = y^{2 }sqrt both sides results in

±x = ± y which MIGHT be true......but not all of the time !

Guest Jan 29, 2021

#5**0 **

This is an excellent question, and I had a similar question when I was first learning. Let's investigate together! Curiosity leads to some interesting results sometimes.

\(x^2=y^2\\ x^2-y^2=0\\ (x-y)(x+y)=0\\ x-y=0\text{ or }x+y=0\\ x=y\text{ or }x=-y\)

With some algebra, we have just proven that x may equal y OR x may equal -y. This should make sense if you think about it.

I will pick an arbitrary value for y, 5.

\(x^2=5^2\\ x^2=25\)

There are two answers to this particular equation, x=5 or x=-5, even though y=5 in both cases. This is exactly what the previous algebra told us, as well.

We can take this question another step further, too! Let's think about \(x^3=y^3\). Does this mean that \(x=y?\). Let's employ the same factoring strategy to see if this is the case.

\(x^3=y^3\\ x^3-y^3=0\\ (x-y)(x^2+xy+y^2)=0\)

Once again, there are two factors. One is an irreducible quadratic, and the other is the familiar (x-y). Let's solve this equation.

\(x-y=0\\ x=y\)

This is a result that we expected. We already know that if you start with the same number that their cube will be equivalent. What is the other mysterious result, you may ask? Well, let's see.

\(x^2+xy+y^2=0\)

This quadratic is not factorable unfortunately, so we will have to employ creative algebra to understand this factor. I will multiply this equation by 2 to achieve

\(2x^2+2xy+2y^2=0\\ {\color{red}{x^2+2xy+y^2}}+x^2+y^2=0\\ {\color{red}{(x+y)^2}}+x^2+y^2=0\)

Notice how the lefthand side of this equation are all quantities that are squared and added together. Squared quantities are always nonnegative, and the only way the addition of nonnegative value can yield 0 is if the quantity that was squared was 0. Therefore, the only possible solution to this equation is that x = 0 and y = 0. That's it! In this particular case x = y, so we already had this situation covered anyway.

Therefore, if \(x^3=y^3\), then \(x=y\). We should note that this property only holds for the real number set.

We can go further once more to \(x^4=y^4\), but I think you understand the idea. I hope this aroused your curiosity!

Guest Jan 29, 2021