How do you express 1 / (x^6 - x^3)1x6−x3 in partial fractions?
1 Answer
Explanation:
First note that:
1/(t^2-t) = 1/(t(t-1)) = (t - (t-1))/(t(t-1)) = 1/(t-1)-1/t1t2−t=1t(t−1)=t−(t−1)t(t−1)=1t−1−1t
So putting
1/(x^6-x^3) = 1/(x^3-1)-1/x^31x6−x3=1x3−1−1x3
We cannot simplify
1/(x^3-1)1x3−1
= 1/((x-1)(x^2+x+1))=1(x−1)(x2+x+1)
= 1/3*3/((x-1)(x^2+x+1))=13⋅3(x−1)(x2+x+1)
= 1/3*((x^2+x+1) - (x^2+x-2))/((x-1)(x^2+x+1))=13⋅(x2+x+1)−(x2+x−2)(x−1)(x2+x+1)
= 1/3*((x^2+x+1) - (x-1)(x+2))/((x-1)(x^2+x+1))=13⋅(x2+x+1)−(x−1)(x+2)(x−1)(x2+x+1)
= 1/3*(1/(x-1)-(x+2)/(x^2+x+1))=13⋅(1x−1−x+2x2+x+1)
= 1/(3(x-1))-(x+2)/(3(x^2+x+1))=13(x−1)−x+23(x2+x+1)
So:
1/(x^6-x^3) = 1/(3(x-1))-(x+2)/(3(x^2+x+1))-1/x^31x6−x3=13(x−1)−x+23(x2+x+1)−1x3
