Question

If a=√2x+3y+√2z+3y/√2x+3y-√2-3y find 3ya^2-4xa+3y​

Answer 1

Answer:

4x^2y + 12xy^2 + 9y^3

Step-by-step explanation:

4x2y + 12xy2 + 9y3

Reorder the terms:

12xy2 + 4x2y + 9y3

Factor out the Greatest Common Factor (GCF), 'y'.

y(12xy + 4x2 + 9y2)

Factor a trinomial.

y((2x + 3y)(2x + 3y))

Final result:

y(2x + 3y)(2x + 3y)

Expression:

4x^2y + 12xy^2 + 9y^3

Answer 2

Answer:

Any value if y=xy=x, 0 if y≠xy≠x

The idea behind is shown below, I will just briefly touch it:

(2x+3y−3x−2y)(2x+3y+3x+2y)=0⇔(y−x)⋅5⋅(x+y)=0(2x+3y−3x−2y)(2x+3y+3x+2y)=0⇔(y−x)⋅5⋅(x+y)=0

So either x=yx=y and x+yx+y can have any arbitrary value or x≠yx≠y and then indeed x+y=0