How do you simplify #x^2/(x^3-125)+5/(x^2+5x+25)#?

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Answer 1 · 2015-02-04T15:19:48

The answer is: #(x^2+5x-25)/((x-5)(x^2+5x+25))#.

First of all, we have to factor all the denominators.

#x^3-125=x^3-5^3=(x-5)(x^2+5x+25)#, with the polynomial #x^2+5x+25# no more factored.

I remember that #a^3-b^3# are called "difference of cubes", and it could be factored:

#a^3-b^3=(a+b)(a^2-ab+b^2)#.

So:

#x^2/((x-5)(x^2+5x+25))+5/(x^2+5x+25)=#

#=(x^2+5(x-5))/((x-5)(x^2+5x+25))=(x^2+5x-25)/((x-5)(x^2+5x+25))#