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anonymous
 5 years ago
x^67x^38=0
anonymous
 5 years ago
x^67x^38=0

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radar
 5 years ago
Best ResponseYou've already chosen the best response.0I 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.

radar
 5 years ago
Best ResponseYou've already chosen the best response.0this 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

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i have that just dont know where to go from there

radar
 5 years ago
Best ResponseYou've already chosen the best response.0\[(x ^{3}=1)=(x+1)(x ^{2}x+1)\]

radar
 5 years ago
Best ResponseYou've already chosen the best response.0That should be (x^3+1) = lol The other factor is the difference of two perfect cubes \[(x ^{3}8)=(x2)(x ^{2}+2x+4)\] Now the total factors can be expressed as:\[(x2)(x+1)(x ^{2}x+1)(x ^{2}+2x+4)\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.07/2 (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

radar
 5 years ago
Best ResponseYou've already chosen the best response.0Don't know how that got in there, seems to be an answer to a different problem

radar
 5 years ago
Best ResponseYou've already chosen the best response.0The web is acting crazy, can't seem to enter anything! Hope you got it all

radar
 5 years ago
Best ResponseYou've already chosen the best response.0I refreshed my browser and it appears ok now. Did you understand the final factors?
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