Calculate the value of M

maxmin

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Subhotosh Khan

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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

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

mmm4444bot

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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

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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....
 
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