Question 697326
A good place to start looking for solutions to cubic functions is to use the Rational Root theorem to try to find rational roots, then use any root found to find a depressed quadratic equation. Any rational roots of a polynomial must be in the form +/- p/q, where p is any number that divides evenly into the constant term, and q is any number that divides evenly into the coefficient of the term with the highest degree. In this case, the constant term is 1 and the coefficient of the x^3 term is also 1, so the only possible rational roots are 1 and -1. Plugging these into the expression yields:<br>

{{{1^3 - 2(1) + 1 = 0}}}, so 1 is a zero.
{{{(-1)^3 - 2(-1) + 1 = 2}}}, so -1 is not a zero.<br>

Since one zero is now known, you can use synthetic division to find a depressed quadratic equation:<br>

<table border="1">
<tr>
<td>1</td>
<td></td>
<td>1</td>
<td>0</td>
<td>-2</td>
<td>1</td>
</tr>
<tr>
<td></td>
<td></td>
<td></td>
<td>1</td>
<td>1</td>
<td>-1</td>
</tr>
<tr>
<td></td>
<td></td>
<td>1</td>
<td>1</td>
<td>-1</td>
<td>0</td>
</tr>
</table><br>

From this, looking at the bottom row of the synthetic division, the depressed equation is {{{x^2 + x - 1 = 0}}}. Use the quadratic formula to solve for x, where a = 1, b = 1, c = -1:<br>

{{{x = (-1 +- sqrt( 1^2-4*1*(-1) ))/(2*1) }}}
{{{x = (-1 +- sqrt(5))/2}}}<br>

So the three zeros are 1 and {{{(-1 +- sqrt(5))/2}}}.