Question

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

Answer 1

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.

Answer 2

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

Answer 3

ok

Answer 4

i have that just dont know where to go from there

Answer 5

\[(x ^{3}=1)=(x+1)(x ^{2}-x+1)\]

Answer 6

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:\[(x-2)(x+1)(x ^{2}-x+1)(x ^{2}+2x+4)\]

Answer 7

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

Answer 8

Don't know how that got in there, seems to be an answer to a different problem

Answer 9

The web is acting crazy, can't seem to enter anything! Hope you got it all

Answer 10

I refreshed my browser and it appears ok now. Did you understand the final factors?