How do you express 2 / (x^3 + 1)2x3+1 in partial fractions?
1 Answer
2/(x^3+1) = 2/(3(x+1))+(2omega)/(3(x+omega))+(2omega^2)/(3(x+omega^2))2x3+1=23(x+1)+2ω3(x+ω)+2ω23(x+ω2)
=2/(3(x+1))+(-2x+4)/(3(x^2-x+1))=23(x+1)+−2x+43(x2−x+1)
where
Explanation:
Let
That is:
omega = -1/2+sqrt(3)/2iω=−12+√32i
Then:
omega^2 = bar(omega) = -1/2-sqrt(3)/2iω2=¯¯ω=−12−√32i
omega^3 = 1ω3=1
1 + omega + omega^2 = 01+ω+ω2=0
x^3+1 = (x+1)(x+omega)(x+omega^2)x3+1=(x+1)(x+ω)(x+ω2)
We find:
1/(x+1)+omega/(x+omega)+omega^2/(x+omega^2)1x+1+ωx+ω+ω2x+ω2
=((x+omega)(x+omega^2)+omega(x+1)(x+omega^2)+omega^2(x+1)(x+omega))/(x^3+1)=(x+ω)(x+ω2)+ω(x+1)(x+ω2)+ω2(x+1)(x+ω)x3+1
=((1+omega+omega^2)x^2+2(1+omega+omega^2)x+3)/(x^3+1)=(1+ω+ω2)x2+2(1+ω+ω2)x+3x3+1
=3/(x^2+1)=3x2+1
Therefore:
2/(x^3+1) = 2/(3(x+1))+(2omega)/(3(x+omega))+(2omega^2)/(3(x+omega^2))2x3+1=23(x+1)+2ω3(x+ω)+2ω23(x+ω2)
If we want a partial fraction decomposition with only Real coefficients, then combine the last two terms so:
=2/(3(x+1))+(2omega(x+omega^2)+2omega^2(x+omega))/(3(x+omega)(x+omega^2))=23(x+1)+2ω(x+ω2)+2ω2(x+ω)3(x+ω)(x+ω2)
=2/(3(x+1))+(-2x+4)/(3(x^2-x+1))=23(x+1)+−2x+43(x2−x+1)
