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manimgoindowndown
Not clear still
I break down the 9x^4 expression using a difference of squares to (3x^2 -2y^2)(3x^2+2y^2) I don't get it from there on especially how the later statment equals 3x^2+2y^2 all by itself

From \((3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2\) divide both parts of the equation by \(3x^2+2y^2\) to get \(3x^2-2y^2=1\) --> \(3x^2=2y^2+1\)--> \(x^2=\frac{2y^2+1}{3}\).
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Ok let's try this one more time since I dont' think you understood which part I didn't get :P

how do you get from/how does (3x^2-2y^2)(2x^2+2y^2) to/equal 3x^2+2y^2
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(3x^2-2y^2)(3x^2+2y^2)=3x^2+2y^2

(3x^2-2y^2)(3x^2+2y^2) - [3x^2+2y^2 ]=0

[3x^2+2y^2 ] [(3x^2-2y^2) - (1) ] = 0
so , either [3x^2+2y^2 ] =0 or [(3x^2-2y^2) - (1) ]=0

[3x^2+2y^2 ] =0 can never be zero as question stem says that x,y are non-zero integers , so 3x^2+2y^2 will always be greater than zero .

[(3x^2-2y^2) - (1) ]=0
3x^2= 2y^2 + 1

x^2= (2y^2 + 1)/3

hence C is the answer.
Hope, it is clear now
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The way I solved it is the following:

9x^4 - 4y^4 = 3x^2 + 2y^2
9x^4 - 3x^2 = 4y^4 + 2y^2
3x^2 ( 3x^2 - 1) = 2y^2 (2y^2 + 1)

3x^2 / 2y^2 = (2y^2 + 1) / (3x^2 - 1)

so now we can equate the numerators for example:

3x^2 = 2y^2 + 1
x^2 = (2y^2 + 1) / 3

Answer: C
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9x^4 – 4y^4 = 3x^2 + 2y^2

Divide both sides by 3x^2 + 2y^2

3x^2 - 2y^2 = 1

x^2 = (2y^2 + 1) / 3 = Answer = C
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9x^4 – 4y^4 = 3x^2 + 2y^2

P^2 - q^2 = (p+q) (p-q)

(3x^2 + 2y^2) (3x^2 - 2y^2 ) = (3x^2 + 2y^2)

as x and y bot non zero integers
(3x^2 + 2y^2) > 0

Divide both sides by 3x^2 + 2y^2

3x^2 - 2y^2 = 1

x^2 = (2y^2 + 1) / 3 = Answer = C
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