Question 381141
<br><font face="Tahoma">We can easily see that {{{f(x)=x^3-1}}} is the difference of two cubes! <br>

{{{f(x)=x^3-1}}}<br>

{{{f(x)=(x-1)(x^2+x+1)}}}<br>

{{{0=(x-1)(x^2+x+1)}}} set f(x) equal to zero, so we have<br>

{{{(x-1)=0}}} or {{{(x^2+x+1)=0}}}<br>

{{{x-1=0}}}<br>

{{{x=1}}} or<br>

{{{(x^2+x+1)=0}}}<br>

This will involve imaginary numbers because there are NO real numbers which satisfy this equation, but we can use the quadratic formula to find the complex solutions involving imaginary numbers:<br>

*[invoke quadratic "x", 1, 1, 1]<br>