question about real numbers My question is:
Solve $(x-a)(x-b)(x-c)=0$  where $a,b,c$ belong to real numbers.
By observing, I found out that $x$ can be $a$ or $b$ or $c$.
 A: In the real numbers, if a product $xy$ is zero (look in your textbook to find this, usually near the "factor theorem"), then one of the terms $x$ or $y$ is zero.  So if $(x-a)(x-b)(x-c)=0,$ then either $x-a=0$, $x-b=0$ or $x-c=0$.  Solve these and you will find $x$ is $a$, $b$ or $c$ respectively.   
A: If $$x_1*x_2*x_3=0，$$
then
      $$x_1=0 \quad\text{ or }\quad x_2=0 \quad\text{ or }\quad x_3=0.$$
In your case,
$$x-a=0 \quad\text{ or }\quad x-b=0 \quad\text{ or }\quad x-c=0.$$
Thus 
$$x=a \quad\text{ or }\quad x=b \quad\text{ or }\quad x=c.$$
A: My interpretation of your question is this: you've found that $x=a,b,c$ solve your equation, and you're curious about whether these are the only solutions to the equation. Suppose that $x$ is not one of $a,b,c$. Then all of the terms $(x-a), (x-b), (x-c)$ are not zero. If you multiply two real numbers that are not zero, then you will end up with something that is not zero. So $(x-a)(x-b)$ is not zero. Similarly, $(x-a)(x-b)(x-c)$ is not zero. This means that any $x$ that is not one of $a$, $b$, or $c$ will not solve your equation.
A: Remember that the real numbers are an integral domain i.e. a commutative ring with no zero divisors. Hence, if we have a product $rs = 0$, this means that $r=0$ or $s=0$. Hence, if you have $$(x-a)(x-b)(x-c) = 0,$$ then either $(x-a)=0$ or $(x-b)=0$ or $(x-c)=0$. Adding $a$, $b$ and $c$ to the three equations respectively gives that $$x =a \text{ or }x =b \text{ or }x =c$$
A: As the other answers mention, your solution is correct. 
The reason for this is the "Zero product property", which applies because the real numbers are a domain (or integral domain).
