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

• MHB
• Monoxdifly
In summary, the equations x^3+3xy^2=a and 3x^2y+y^3=b can be solved using complex numbers. The solutions for x and y are x = Re(a+ib)^1/3 and y = Im(a+ib)^1/3. If the signs were reversed on the second equation, the solutions would be x+y = (a+b)^1/3 or x-y = (a-b)^1/3.
Monoxdifly
MHB
$$\displaystyle x^3-3xy^2=a$$
$$\displaystyle 3x^2y-y^3=b$$
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?

Monoxdifly said:
$$\displaystyle x^3-3xy^2=a$$
$$\displaystyle 3x^2y-y^3=b$$
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 $y$.

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 ...

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

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

skeeter said:
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 $$\displaystyle x+y=\sqrt[3]{a+b}$$.

... could also be $x-y=\sqrt[3]{a-b}$

Ugh... If only that$$\displaystyle 3xy^2$$ on the question was positive, this would be easy to answer.

## 1. What is the relationship between x and y in the given equations?

The given equations show a relationship between x and y in the form of a cubic equation and a quadratic equation. This relationship can be further explored by solving for either x or y and graphing the resulting equation.

## 2. How do I make x the subject in the first equation?

To make x the subject in the first equation, we can rearrange the equation to isolate x. This can be done by dividing both sides of the equation by 3y^2 and then taking the cube root of both sides.

## 3. How do I make y the subject in the second equation?

To make y the subject in the second equation, we can rearrange the equation to isolate y. This can be done by dividing both sides of the equation by 3x^2 and then taking the cube root of both sides.

## 4. How can I solve for x and y simultaneously?

To solve for x and y simultaneously, we can substitute the expression for x from the first equation into the second equation. This will result in a single equation with only y as the variable, which can then be solved for y. Once y is known, we can substitute this value back into the first equation to solve for x.

## 5. What is the significance of the constants a and b in the equations?

The constants a and b represent the coefficients in the given equations and are used to define the relationship between x and y. Depending on their values, the equations can have different solutions and can represent different curves when graphed.

• General Math
Replies
4
Views
812
• General Math
Replies
7
Views
1K
• General Math
Replies
18
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
532
• General Math
Replies
2
Views
2K
• Precalculus Mathematics Homework Help
Replies
11
Views
846
• General Math
Replies
2
Views
1K
• General Math
Replies
1
Views
867
• General Math
Replies
3
Views
1K
• Calculus and Beyond Homework Help
Replies
24
Views
2K