#### Answer

$x=1,\pm \sqrt{3}$

#### Work Step by Step

$f(x)=x^3-x^2-3x+3$
Since we are solving for all the real values of $x$ when $f(x) = 0$, set the function $f(x)=x^3-x^2-3x+3$ equal to zero and we get the following equation:
$x^3-x^2-3x+3=0$
Notice that this needs to be solved using factoring by grouping. Start by grouping the left two terms and the right two terms.
$(x^3-x^2)+(-3x+3)=0$
Then, factor as much out of the left two terms as possible. In this example, factor out $x^2$:
$x^2(x-1)+(-3x+3)=0$
Next, factor the $(x-1)$ term out of the right two terms:
$x^2(x-1)+(x-1)(-3)=0$
Now, notice that you can factor an $(x-1)$ term out of the entire left side of the equation:
$(x-1)(x^2-3)=0$
Next, you can see that the above equation holds true when $x-1=0$ or $x^2-3=0$. Solve for $x$ in each equation individually as shown below:
$x-1=0$
Add 1 to both sides.
$x-1+1=0+1$
$x=1$
Now for the other equation:
$x^2-3=0$
Add 3 to both sides of the equation:
$x^2-3+3=0+3$
$x^2=3$
Now solve for x by taking the square root of both sides.
$\sqrt{x^2}=\sqrt{3}$
Remember that taking the square root of both sides means we need to include the $\pm$ on the right side of the equation.
$x=\pm \sqrt{3}$
So, the final solution is $x=1,\pm \sqrt{3}$