# algebra equations

• Dec 8th 2008, 11:43 PM
Twertz
algebra equations
x+y=xy=3

find x^3 + y^3

Anyone? Could really use some help on this the extra credit would be nice to have.
The work is more important than the answer, so any idea on how to get started so i could begin figuring it would be much appreciated.
• Dec 8th 2008, 11:57 PM
Chop Suey
Quote:

Originally Posted by Twertz
The work is more important than the answer, so any idea on how to get started so i could begin figuring it would be much appreciated.

I'm not going to solve it to give you a chance. Here's my hint:
$(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3$

Try and come back if you run into trouble.
• Dec 9th 2008, 12:05 AM
Twertz
ok i'm still stuck. it's Find x^3 + y^3 so am i trying to cube the original equation? a bit more of a boost maybe?

i've got to be taking it in the wrong direction

i cubed the original equation and moved everything so it was x^3 + y^3= -3x^2y-3xy^2-x^3y^3-27
• Dec 9th 2008, 12:20 AM
Chop Suey
Quote:

Originally Posted by Twertz
ok i'm still stuck. it's Find x^3 + y^3 so am i trying to cube the original equation? a bit more of a boost maybe?

i've got to be taking it in the wrong direction

i cubed the original equation and moved everything so it was x^3 + y^3= -3x^2y-3xy^2-x^3y^3-27

Well, I'm mystified on how x^3y^3 came to be. Your manipulation is incorrect.

$(x+y)^3 = x^3 + 3x^2y+3xy^2+y^3 \implies x^3 + 3xy(x+y) + y^3 = (x+y)^3$

Can you do it now?
• Dec 9th 2008, 12:24 AM
nzmathman
You have $x + y = 3$

So $(x + y)^3 = 27$

$(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

$27 = x^3 + 3x^2y + 3xy^2 + y^3$

$x^3 + y^3 = 27 - 3x^2y - 3xy^2$

$x^3 + y^3 = 27 - 3xy(x + y)$

$x^3 + y^3 = 27 - 3 \times 3 \times 3 \quad$ (since x+y = 3 and xy = 3
• Dec 9th 2008, 01:06 AM
mr fantastic
Quote:

Originally Posted by nzmathman
You have $x + y = 3$

So $(x + y)^3 = 27$

$(x + y)^3 = x^3 + 3x^2y + 3xy^2 + y^3$

$27 = x^3 + 3x^2y + 3xy^2 + y^3$

$x^3 + y^3 = 27 - 3x^2y - 3xy^2$

$x^3 + y^3 = 27 - 3xy(x + y)$

$x^3 + y^3 = 27 - 3 \times 3 \times 3 \quad$ (since x+y = 3 and xy = 3

You do realise that Chop Suey was attempting to guide the OP to an answer ..... By posting the complete solution you have completely undermined his efforts.
• Dec 9th 2008, 02:24 AM
CaptainBlack
Quote:

Originally Posted by mr fantastic
You do realise that Chop Suey was attempting to guide the OP to an answer ..... By posting the complete solution you have completely undermined his efforts.

Not only what MrF says, but this is an extra credit assignment, so if the OP sees this and uses it he is cheating (extra credit = part of assessment).

CB