Question 260277
Here is the original question:
{{{x^3+729=0 }}}
we can express this as the sum of cubes, but first write them both as powers of 3, as
{{{x^3+9^3 = 0}}}
Now, we get
{{{(x+9)(x^2-9x+81)=0}}}
next set each part =0 and solve.
x+9 = 0
x = -9
--
x^2-9x+81= 0 
by quadratic, we get
{{{x = (9 +- sqrt( 81-4*1*81 ))/(2) }}}
which simplifies to
{{{x = (9 +- sqrt(81-324))/(2) }}}
which is
{{{x = (9 +- sqrt(-243 ))/(2) }}}
which will give us imaginary roots as
{{{x = (9 +- 9*isqrt(3))/2}}}
--
we have three roots
x = -9
x = (9+9isqrt(3))/2
x = (9-9isqrt(3))/2