SOLUTION: x^6 - 511x^3-512 = 0 find all real solutions of the equation. hi, im having trouble factoring this out..getting confused with the exponents. thank you for your time!

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Question 498680: x^6 - 511x^3-512 = 0
find all real solutions of the equation.
hi, im having trouble factoring this out..getting confused with the exponents.
thank you for your time!
Found 2 solutions by richard1234, Earlsdon:
Answer by richard1234(7193) About Me  (Show Source):
You can put this solution on YOUR website!
We can let y = x^3, so our polynomial becomes

y^2 - 511y^2 - 512 = 0, which factors to

(y-512)(y+1) = 0

y = 512, -1 = x^3

The only real solutions are x = -1 and x = 8.

Answer by Earlsdon(6294) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E6-511x%5E3-512+=+0 Treat this as a quadratic equation in x%5E3 thus:
%28x%5E3%29%5E2-511%28x%5E3%29-512+=+0 Now factor this.
%28x%5E3%2B1%29%28x%5E3-512%29+=+0 Apply the "zero product" rule:
x%5E3%2B1+=+0 or x%5E3-512+=+0
Now since the original equation is a 6th order equation, we can expect 6 roots.
Here's how we get them:
x%5E3%2B1+=+0 This is the sum of cubes which can be factored thus:
x%5E3%2B1+=+%28x%2B1%29%28x%5E2-x%2B1%29so...
%28x%2B1%29%28x%5E2%2Bx%2B1%29+=+0 Apply the "zero product" rule:
x%2B1+=+0 so highlight%28x+=+-1%29
x%5E2%2Bx%2B1+=+0 Apply the quadratic formula to get:
highlight%28x+=+5%2B0.866i%29 and highlight%28x+=+5-0.866i%29
For the other three roots, we have:
x%5E3-512+=+0 This is the difference of cubes %28x%29%5E3-%288%29%5E3which factors as:
%28x-8%29%28x%5E2%2B8x%2B64%29+=+0 Apply the "zero product" rule:
x-8+=+0 so highlight%28x+=+8%29
x%5E2%2B8x%2B64+=+0 Apply the quadratic formula to get:
highlight%28x+=+-4%2B6.928i%29 and highlight%28x+=+-4-6.928i%29