Question

pls help me..

Answer 1

post your question, We'll try to help

Answer 2

Are you trying to simplify the equation?

Answer 3

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

Answer 4

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

Answer 5

just check and tell

Answer 6

(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))

Answer 7

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

Answer 8

its a fraction

Answer 9

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
\[(x^2+2bx-ax-2ab) b(x+2b^2)(3x+8y)(9x^2-6xy+4y^2)\]

Answer 10

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
\[\frac{y(3x^2-39x+2xy-2ay)(x-2b)}{(9x^2+2bx-ax-2ab)(3x+8y)}\]