What is $\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$?

Question

What's the result of: $$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}$$ Is it $$\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|x|}=0$$ or $$\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}=\frac{1}{|{x}|}-\frac{1}{x}\frac{x^2}{|x|^2}=\frac{1}{|{x}|}-\frac{1}{x}=\left\{\begin{matrix}\frac{2}{x},x<0\\ 0,x>0 \end{matrix}\right.$$

Answer 1

As said in the comments, you have $|x|^2=x^2$ which is why the first proof is correct.

The error in the second proof is in the very beginning, namely $\frac{1}{|{x}|}-\frac{x^2}{|x|^2x}$ which is different than what you started with. It is wrong that $|x|^3=|x|^2x$. Take $x=-1$ for example.

Answer 2

Yes the first one is correct, indeed we have $|x|^3=|x||x^2|=|x|x^2$ then

$$\frac{1}{|{x}|}-\frac{x^2}{|x|^3}=\frac{1}{|{x}|}-\frac{x^2}{|x|x^2}=\frac{1}{|{x}|}-\frac{1}{|{x}|}=0$$

For the same reason the second one is wrong.

Answer 3

Your first conclusion is right since $$|x|^3=|x|^2\cdot |x|=x^2\cdot |x|$$and the second is wrong since for $x<0$ $$|x|^3=-x^3\ne x^3= |x|^2\cdot x$$

Answer 4

As the other answers already told you, the first one is true.

But maybe this way of thinking gives you an intuition: Both terms $\frac{1}{|x|}$ and $\frac{x²}{|x³|}$ are positive and obviously have the same absolute value, so since there is a minus in between, the result must be zero.