Algebra problem?

[SilverSprite]

If x,y,z are part of the set of reals where x <> y <> z, whose sum is zero and product is 2, find values of x^3 +y^3 + z^3.

[SilverSprite]

Geez how come i got -6?

[SilverSprite]

so you didnt actually do it?

[SilverSprite]

fine then and i think 91 was the answer to the one with the three roots right?

[DiGiTaIErRoR]

Originally posted by SilverSprite
Geez how come i got -6?

Prove it.

[SilverSprite]

x+y+z=0
xyz=2

(x+y+z)^3=0^3

Expand and get:
x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3zx^2+3yz^2+3zy^2+6xyz=0

Sub in xyz=2 to get:
x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3zx^2+3yz^2+3zy^2=12

Common factor:
x^3+y^3+z^3+3x^2 (y+z)+3y^2 (x+z)+3z^2 (x+y)=12

Sub solving for x,y,z in the first statment and subbing them back in then subtracting:
- 2x^3 - 2y^3 - 2z^3=12

Dividing both sides by -2 you get:
x^3+y^3+z^3=-6

[SilverSprite]

Or that would work. And your right. And i'm not sure how you did it though.

[Lior]

Hey Silver, is it nice talking to yourself?

You might suffer a personality split.





...Oh, and you're too

[DiGiTalErRoR]

Excuse me.

x^3 +y^3 + z^3 != (x+y+z)^3

(x+y+z)^3=(x+y+z)(x+y+z)(x+y+z) != x^3 +y^3 + z^3

[DiGiTalErRoR]

Unless I'm not understanding the original question.

[DiGiTalErRoR]

When I put in x^3+y^3+z^3 and x+y+z=0 and x*y*z=2 into my ti-92, it returns false.

And if x*y*z=2 then how can you get a negative result for them ^3? The signs would be the same. Thus the result would be a positive.

[DiGiTalErRoR]

Wait. I know

-.9999... recurring, -1, 2

I get a positive 6.

[Lior]

Yo Digital, I see you made a new nick.

Congratulations.

[NotLKH]

Originally posted by SilverSprite
x+y+z=0
xyz=2

(x+y+z)^3=0^3

Expand and get:
x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3zx^2+3yz^2+3zy^2+6xyz=0

Sub in xyz=2 to get:
x^3+y^3+z^3+3xy^2+3xz^2+3yx^2+3zx^2+3yz^2+3zy^2=-12

Common factor:
x^3+y^3+z^3+3x^2 (y+z)+3y^2 (x+z)+3z^2 (x+y)=-12

Sub solving for x,y,z in the first statment and subbing them back in then subtracting:
- 2x^3 - 2y^3 - 2z^3=-12

Dividing both sides by -2 you get:
x^3+y^3+z^3=6

Sorry, I saw a small error involving the sub of xyz=2 that I Just had to fix.

-Lou

[SilverSprite]

My mistake

[bugzpodder]

what happened to the really nice proof that got the answer in three lines?