# Calculate the value of M

#### Subhotosh Khan

##### Super Moderator
Staff member
What are 'x' & 'y'? Where are 'x' & 'y'?

$$\displaystyle \displaystyle{Calculate \ \ expressions \ \ value \ \ M \ \ = \ \ a^2 \ \ + \ \ b^2 \ \ know \ \ x \ \ and \ \ y \ \ satisfy:}$$

$$\displaystyle \displaystyle{\frac{3b^2}{a^2} + \frac{2}{a^3} \ = 1}$$ and

$$\displaystyle \displaystyle{\frac{3a^2}{b^2} + \frac{1}{b^3} \ = 1}$$

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#### tkhunny

##### Moderator
Staff member
Perhaps you could share the ENTIRE problem and show ANY of your own work? That might give us a clue.

#### mmm4444bot

##### Super Moderator
Staff member
In this case, your exercise involves solving a system of two equations.

$$\displaystyle \dfrac{3y^2}{x^2} + \dfrac{2}{x^3} = 1$$

$$\displaystyle \dfrac{3x^2}{y^2} + \dfrac{1}{y^3} = 1$$

Both of these are cubic equations. Solving this system by hand seems too complex, for me.

I used software to graph the functions; there are three Real solutions. Wolfram-Alpha reports five additional solutions, each containing an imaginary part.

#### Subhotosh Khan

##### Super Moderator
Staff member
In this case, your exercise involves solving a system of two equations.

$$\displaystyle \dfrac{3y^2}{x^2} + \dfrac{2}{x^3} = 1$$

$$\displaystyle \dfrac{3x^2}{y^2} + \dfrac{1}{y^3} = 1$$

Both of these are cubic equations. Solving this system by hand seems too complex, for me.

I used software to graph the functions; there are three Real solutions. Wolfram-Alpha reports five additional solutions, each containing an imaginary part.
Assuming $$\displaystyle x \ne 0$$ & $$\displaystyle y \ne 0$$

3y2x + 2 = x3

3x2y + 1 = y3

Subtracting:

(x-y)(x2 +xy + y2) + 3xy(x-y) = 1

(x-y)(x2 + 4xy + y2) = 1

Adding:

(x+y)(x2 - xy + y2) - 3xy(x+y) = 3

(x+y)(x2 - 4xy + y2) = 3

x2 + y2 = 1/2 * [1/(x-y) + 3/(x+y)]

I have not gone any further....