divide and simplify[( 3x-x^2)/(x^3-27)] divide by [ (x)/(x^2+3x+9)]

Question

[( 3x-x^2)/(x^3-27)] divide by [ (x)/(x^2+3x+9)]

Answer 1

x^3-27 factorises, because x=3 is clearly a root when x^3-27=0.
If we divide by the factor (x-3) the other factor is x^2+3x+9.
3x-x^2 also factorises: x(3-x). So (3x-x^2)/(x^3-27) simplifies to x(3-x)/((x-3)(x^2+3x+9))=-x/(x^2+3x+9).
Dividing further by x/(x^2+3x+9), which is the same as multiplying by (x^2+3x+9)/x, gives us -1.

Answer 2

Given [(3x-x^2)/(x^3-27)] / [x/(x^2+3x+9)] =[(3x-x^2)/(x^3-27)] *(x^2+3x+9)/x] but x^3-27 = (x-3)(x^2+3x+9) =[(3x-x^2)/(x-3)(x^2+3x+9)] * [(x^2+3x+9)/x] =(3x-x^2)/[x(x-3)] = -x(3-x)/[x(x-3)] = -1