## anonymous 4 years ago pls help me..

1. ash2326

post your question, We'll try to help

2. anonymous

Are you trying to simplify the equation?

3. anonymous

What are you trying to do here? Are x and y independent variables and a and b constants?

4. ash2326

Zhenecca I think the first term should be (9x^2y-6xy+4y^3) it should not be -6xy2

5. ash2326

just check and tell

6. anonymous

(9x^2y-6xy2+4y^3)/(x^2+2bx-ax-2ab)÷((bx+2b^2)/(3x^2-39x+2xy-2ay)*(27x^3+8y^3)/(bx^2-4b^3))

7. ash2326

as the function is very long, I'll try to simplify each term one by one

8. anonymous

its a fraction

9. ash2326

We'll have in the numerator $(9x^2y-6xy^2+4y^3)(3x^2-39x+2xy-2ay)(bx^2-4b^3)$ in the denominator we'll have $(x^2+2bx-ax-2ab)(bx+2b^2)(27x^3+8y^3)$ now the numerator can be written as $y(9x^2-6xy+4y^2)(3x^2-39x+2xy-2ay) b(x^2-4b^2)$ denominator can be written as $(x^2+2bx-ax-2ab) b(x+2b^2)(3x+8y)(9x^2-6xy+4y^2)$

10. ash2326

now canceling common terms from numerator and denominator numerator will be $y(3x^2-39x+2xy-2ay)b(x-2b)$ and denominator will be $(9x^2+2bx-ax-2ab)b(3x+8y)$ so we get finally $\frac{y(3x^2-39x+2xy-2ay)(x-2b)}{(9x^2+2bx-ax-2ab)(3x+8y)}$