Question
Assume that no denominator equals 0.
(8y^3+27)/(2xy-10y+3x-15)
I don't know how to solve this one.
(8y^3+27)/(2xy-10y+3x-15)
I don't know how to solve this one.
Answers
Reiny
You will have to factor both top and bottom and then reduce
your text should have a rule for the sum of cubes
(A^3 + B^3) = (A+B)(A^2 - AB + b^2)
so the top factors to
(2y+3)(4y^2 - 6y + 9)
the bottom factors by grouping to
(x-5)(2y+3)
so after you cancel you would have
(x-5)/(4y^2 - 6y + 9)
your text should have a rule for the sum of cubes
(A^3 + B^3) = (A+B)(A^2 - AB + b^2)
so the top factors to
(2y+3)(4y^2 - 6y + 9)
the bottom factors by grouping to
(x-5)(2y+3)
so after you cancel you would have
(x-5)/(4y^2 - 6y + 9)