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

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Discussion Overview

The discussion revolves around the mathematical problem of expressing the variables x and y in terms of the parameters a and b, given the equations \(x^3 - 3xy^2 = a\) and \(3x^2y - y^3 = b\). The scope includes mathematical reasoning and exploration of potential solutions, including the use of complex numbers.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants suggest manipulating the original equations to combine them, leading to the expression \((x-y)(x^2+4xy+y^2) = a+b\), but express uncertainty about the next steps.
  • One participant proposes that the problem can be approached using complex numbers, leading to the formulation \((x+iy)^3 = a + ib\) and deriving solutions for x and y as the real and imaginary parts of \((a+ib)^{1/3}\).
  • Another participant comments that the complexity of the problem suggests it may not be suitable for high-school level understanding.
  • Some participants propose alternative equations, such as \(x^3 + 3xy^2 = a\) and \(3x^2y + y^3 = b\), arguing that these would be more appropriate for the forum's audience.
  • One participant expresses frustration about the specific term \(3xy^2\) being negative, suggesting it complicates finding a straightforward solution.

Areas of Agreement / Disagreement

Participants express differing views on the complexity of the original problem and its suitability for the forum. There is no consensus on a definitive method for solving the equations, and multiple approaches are discussed without resolution.

Contextual Notes

Some participants note that the original equations may lead to complex solutions, while others suggest simpler forms that could yield more accessible answers. The discussion reflects varying levels of mathematical background among participants.

Who May Find This Useful

Readers interested in advanced algebra, complex numbers, and problem-solving strategies in mathematics may find this discussion relevant.

Monoxdifly
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$$x^3-3xy^2=a$$
$$3x^2y-y^3=b$$
make a formula using a and b for x and y.

What I've done:
$$x^3-3xy^2+3x^2y-y^3=a+b$$
$$x^3+3x^2y-3xy^2-y^3=a+b$$
$$(x-y)(x^2+4xy+y^2)=a+b$$
I don't know what to do from here. Any hints?
 
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Monoxdifly said:
$$x^3-3xy^2=a$$
$$3x^2y-y^3=b$$
make a formula using a and b for x and y.

What I've done:
$$x^3-3xy^2+3x^2y-y^3=a+b$$
$$x^3+3x^2y-3xy^2-y^3=a+b$$
$$(x-y)(x^2+4xy+y^2)=a+b$$
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 $y$.
 
Opalg said:
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.
 
If the equations were ...

$$x^3+3xy^2=a$$
$$3x^2y+y^3=b$$

... then the problem would fit the level of this forum.
 
skeeter said:
If the equations were ...

$$x^3+3xy^2=a$$
$$3x^2y+y^3=b$$

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

Guess I better change the question to that and make the answer $$x+y=\sqrt[3]{a+b}$$.
 
... could also be $x-y=\sqrt[3]{a-b}$
 
Ugh... If only that$$3xy^2$$ on the question was positive, this would be easy to answer.
 

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