x^6-7x^3-8=0

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At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

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I am assuming that you want to simplify this. It will simplify first factor it like this: \[(x ^{3}+1)(x ^{3}-8)\] The second factor is the difference of two perfect cubes and can be factored further.
this can be factored further as one is the sum of two perfect cubes, and the other is the difference. I will do them one at a time and then combine
ok

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i have that just dont know where to go from there
\[(x ^{3}=1)=(x+1)(x ^{2}-x+1)\]
That should be (x^3+1) = lol The other factor is the difference of two perfect cubes \[(x ^{3}-8)=(x-2)(x ^{2}+2x+4)\] Now the total factors can be expressed as:\[(x-2)(x+1)(x ^{2}-x+1)(x ^{2}+2x+4)\]
7/2 (7/2)^2=49/4 x^6-(7/2)^3=8+49/4 (x^3-(7/2))^2=81/4 x^3-(7/2)=+ or - sqrt{81/4} x^3=(7/2) + or - sqrt{81/4} x^3=(7/2) + or - sqrt{(81*4)/4} x^3=(7/2) + or - sqrt{324} x^3=(7/2) + or - 9/2 x^3=(7/2)+(9/2) x^3=8 x=2^3 x=2 or: x^3=(7/2)-(9/2) x^3=-2/2 x=sqrt[3]{-1} x=-1
Don't know how that got in there, seems to be an answer to a different problem
The web is acting crazy, can't seem to enter anything! Hope you got it all
I refreshed my browser and it appears ok now. Did you understand the final factors?

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