partial frac.

[hrr379]

Please let me know if I have this done correctly?

(t^6 + 1)/(t^6 +t^3)

solution: (t^6 + 1)/(t^3(t^3+1)
A/t +B/(t^2) + C/(t^3) + D/(t^3 +1)

 

[tkhunny]

Numerator Degree = Denominator Degree

This calls for a division problem, before doing anything else. Alternatively, you could include a partial denominator of '1'.

 

[soroban]

Hello, hrr379!

\(\displaystyle \dfrac{t^6 + 1}{t^6 +t^3}\)

As tkhunny suggested, use long division:
. . \(\displaystyle \dfrac{t^6 + 1}{t^6 + t^3} \;=\;1 - \dfrac{t^3-1}{t^6 + t^3}\)

Then apply Partial Fractions to that fraction:
. . \(\displaystyle \dfrac{t^3-1}{t^3(t^3+1)} \;=\;\dfrac{t^3-1}{t^3(t+1)(t^2-t+1)} \)

We have: .\(\displaystyle \displaystyle\frac{t^3-1}{t^3(t+1)(t^2-t+1)} \;=\;\frac{A}{t} + \frac{B}{t^2} + \frac{C}{t^3} + \frac{D}{t+1} + \frac{Et+F}{t^2-t+1}\)

Good luck!