$x(x^2-2)=0$, The answers are $x = 0, \sqrt{2}$, how do I get there? $$x(x^2-2)=0$$
The answers are $x=\sqrt{2}, 0$ how do I get there?

Let's start from the beginning.
You are given

$$x(x^2 - 2 ) = 0$$

What are they asking for here?
Well, they're asking for any value of $x$ which would cause the given statement to become true.
What does this mean?
It means that if you were to substitute a number in place of $x$, then solve out the equation, the resulting statement must make logical sense.
How do we find these values of $x$?
We can make a few observations that will aide us in the discovery of these values. For instance, we could separate the equation into two parts then solve each individually.
Observe,
$$x(x^2-2) = 0$$
$$x\qquad(x^2-2) = 0$$
$$x=0,\qquad(x^2-2) = 0$$
Why do we do this?
Well, we want to find a value for $x$ that makes the whole equation true, and since the equation is just a product, if a $0$ is the result anywhere in the equation the entire product will become $0$.
Less talk, more proof.
$$
\begin{align}
x(x^2-2) &= 0\qquad x(x^2-2) = 0\\
(0)(x^2-2) &= 0\qquad\qquad\; x(0) = 0\\
0 &= 0\qquad\qquad\quad\ \ \  0 = 0
\end{align}
$$
$$\color{green}{True!}$$
So , we can break up this equation into two parts and just solve each individually for an $x$ that will be equal to $0$.
Now let's start actually doing this.
We have already divided up the equation above
$$x=0,\qquad(x^2-2) = 0$$
As you can see, we already have a value for $x$ given just by dividing up the equation, that value being $x=0$. Now, we have to see if there are any other possible solutions that can be given by solving the other part of the equation.
Let's look at
$$(x^2-2) = 0$$
All we have to do here is a bit of algebra to solve for $x$.
$$
\begin{align}
x^2-2 & = 0\\
x^2-2+2 & = 0+2\\
x^2 & = 2\\
\sqrt{x^2} & = \pm\sqrt{2}\\
x & = \pm\sqrt{2}\\
\end{align}
$$

Why is there a $\pm$ in front of $\sqrt{2}$?
That is a 'plus or minus' sign, it literally means that the following value can be either positive or negative.
Why do we need it?
Well, squaring a real number will  always produce a positive number, regardless of whether the real number being squared is positive or negative.
$$(2)^2 = 4$$
$$(-2)^2 = 4$$
Therefore, when we are undoing a square, in other words taking a square root, we cannot assume that the result is just positive or negative, since it could be either and still be true. So, we have to split the result of a square root up into two components, its positive result and its negative result, and solve the resulting equations using each result in turn.
In summary,
$$
\begin{align}
x &= \pm\sqrt{2}\\
\implies x=\sqrt{2}&,\quad x=-\sqrt{2}\\
\end{align}
$$

Now that we have our values for $x$, we have to check them to make sure nothing was done incorrectly.
$$\color{red}{\text{Always check your work!}}$$
Given:
$$x(x^2 - 2 ) = 0$$
Tests:
$$x=0\qquad x=\pm\sqrt{2}$$
We see:
$$
\begin{align}
x(x^2 - 2 ) &= 0 \qquad\qquad\quad x(x^2 - 2 ) = 0 \qquad\qquad\qquad\ \ \  x(x^2 - 2 ) = 0\\
(0)((0)^2 - 2 ) &= 0 \qquad (\sqrt{2})((\sqrt{2})^2 - 2 ) = 0 \qquad (-\sqrt{2})((-\sqrt{2})^2 - 2 ) = 0\\
(0)(0-2) &= 0 \qquad\qquad (\sqrt{2})(2 - 2) = 0 \qquad\qquad\quad (-\sqrt{2})(2 - 2 ) = 0\\
(0)(-2) &= 0 \qquad\qquad\quad\ \ \  (\sqrt{2})(0) = 0 \qquad\qquad\qquad\ \ \  (-\sqrt{2})(0) = 0\\
0 &= 0 \qquad\qquad\qquad\qquad\ \ \   0 = 0 \qquad\qquad\qquad\qquad\qquad\ \ 0 = 0\\
\end{align}
$$
$$
\color{green}{True!}
$$
Everything works out.
Tips hat, walks away.
