SOLUTION: Solve this equation: x^3+729=0 I know this much: 1st solution is -9 (x+9)^3=0 (x+9)[x^2-9x+81] This is where I'm stuck..When I use the quadratic formula I get 9 +/-squa

Question 260277: Solve this equation: x^3+729=0
I know this much: 1st solution is -9
(x+9)^3=0
(x+9)[x^2-9x+81]
This is where I'm stuck..When I use the quadratic formula I get
9 +/-square root of 243 over 2
Now what? The answer is 9 +9i 'square root' of 3 over 2. How do you get that?
Answer by drk(1908) (Show Source): You can put this solution on YOUR website!
Here is the original question:

we can express this as the sum of cubes, but first write them both as powers of 3, as

Now, we get

next set each part =0 and solve.
x+9 = 0
x = -9
--
x^2-9x+81= 0
by quadratic, we get

which simplifies to

which is

which will give us imaginary roots as

--
we have three roots
x = -9
x = (9+9isqrt(3))/2
x = (9-9isqrt(3))/2
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