Solve:

$\displaystyle x^3-y^3=602$

$\displaystyle xy(y-x)=-198$

Answer:

x = 11 and y = 9

or

x = -9 and y = -11

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- Dec 12th 2008, 09:36 AMApprentice123System
Solve:

$\displaystyle x^3-y^3=602$

$\displaystyle xy(y-x)=-198$

Answer:

x = 11 and y = 9

or

x = -9 and y = -11 - Dec 12th 2008, 10:00 AMChop Suey
I will give you two hints:

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

$\displaystyle xy^2 - x^2y=-198 \implies -3x^2y+3xy^2=-594$ - Dec 12th 2008, 10:14 AMApprentice123
How i find the answer ?

- Dec 12th 2008, 10:20 AMChop Suey
$\displaystyle x^3-y^3=602$

$\displaystyle -3x^2y+3xy^2=-594$

If we combine the two expressions, we get a perfect cube:

$\displaystyle x^3 - 3x^2y+3xy^2-y^3=602-594 \implies (x-y)^3=8 \implies x - y =\ldots$

This is your key to solving for x and y. Can you do it now? - Dec 12th 2008, 10:36 AMApprentice123
thank you