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[ASK] Make x and y the Subjects: x^3-3xy^2=a, 3x^2y-y^3=b

Monoxdifly

Well-known member
Aug 6, 2015
267
\(\displaystyle x^3-3xy^2=a\)
\(\displaystyle 3x^2y-y^3=b\)
make a formula using a and b for x and y.

What I've done:
\(\displaystyle x^3-3xy^2+3x^2y-y^3=a+b\)
\(\displaystyle x^3+3x^2y-3xy^2-y^3=a+b\)
\(\displaystyle (x-y)(x^2+4xy+y^2)=a+b\)
I don't know what to do from here. Any hints?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,656
Leeds, UK
\(\displaystyle x^3-3xy^2=a\)
\(\displaystyle 3x^2y-y^3=b\)
make a formula using a and b for x and y.

What I've done:
\(\displaystyle x^3-3xy^2+3x^2y-y^3=a+b\)
\(\displaystyle x^3+3x^2y-3xy^2-y^3=a+b\)
\(\displaystyle (x-y)(x^2+4xy+y^2)=a+b\)
I don't know what to do from here. Any hints?
It looks to me as though this question is designed to be tackled using complex numbers. In fact, if you add $i$ times the second equation to the first equation, you get $x^3-3xy^2 + i(3x^2y-y^3) = a + ib$, which can be written $(x+iy)^3 = a + ib$. So $x+iy = (a+ib)^{1/3}$, and the solutions for $x$ and $y$ are then $$x = \operatorname{Re}(a+ib)^{1/3}, \qquad y = \operatorname{Im}(a+ib)^{1/3}.$$ Since a complex number has three cube roots, those formulas give three solutions for $x$ and $y$.
 

Monoxdifly

Well-known member
Aug 6, 2015
267
It looks to me as though this question is designed to be tackled using complex numbers. In fact, if you add $i$ times the second equation to the first equation, you get $x^3-3xy^2 + i(3x^2y-y^3) = a + ib$, which can be written $(x+iy)^3 = a + ib$. So $x+iy = (a+ib)^{1/3}$, and the solutions for $x$ and $y$ are then $$x = \operatorname{Re}(a+ib)^{1/3}, \qquad y = \operatorname{Im}(a+ib)^{1/3}.$$ Since a complex number has three cube roots, those formulas give three solutions for $x$ and $y$.
Complex number, eh? So, it's definitely not for high-schoolers, then.
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
If the equations were ...

\(\displaystyle x^3+3xy^2=a\)
\(\displaystyle 3x^2y+y^3=b\)

... then the problem would fit the level of this forum.
 

Monoxdifly

Well-known member
Aug 6, 2015
267
If the equations were ...

\(\displaystyle x^3+3xy^2=a\)
\(\displaystyle 3x^2y+y^3=b\)

... then the problem would fit the level of this forum.
Guess I better change the question to that and make the answer \(\displaystyle x+y=\sqrt[3]{a+b}\).
 

skeeter

Well-known member
MHB Math Helper
Mar 1, 2012
617
North Texas
... could also be $x-y=\sqrt[3]{a-b}$
 

Monoxdifly

Well-known member
Aug 6, 2015
267
Ugh... If only that\(\displaystyle 3xy^2\) on the question was positive, this would be easy to answer.