Question

Convert and resolve into partial fraction x^3+1 / x^2+6x​

Answer 1

Answer:

You need a field extension. Let a^3+a^2+1=0, then you can factor

x^3+x^2+1=(x-a)*(x^2+a*x+x+a^2+a) and anytime you see a^3 or higher, use

a^3=-1-a^2 to reduce the power. It doesn’t matter which root you take, a^3=-1-a^2 holds true.

Then 1/(x^3+x^2+1)=1/((x-a)*(x^2+a*x+x+a^2+a))=

k1/(x-a)+(k2*x+k3)/(x^2+a*x+x+a^2+a) so multiply by

x^3+x^2+1=(x-a)*(x^2+a*x+x+a^2+a) on both sides to get

1=k1*(x^2+a*x+x+a^2+a)+(k2*x+k3)*(x-a)=

x^2*(k1+k2)+x*(k1*a+k1-k2*a+k3)+k1*a^2+k1*a-k3*a so equate powers of x to get

x^2: 0=k1+k2 so k2=-k1

x: 0=k1*a+k1-k2*a+k3=k1*a+k1+k1*a+k3 so k3=-k1*(1+2*a)

x^0: 1=k1*a^2+k1*a-k3*a=

k1*a^2+k1*a-(-k1*(1+2*a))*a=k1*a^2+k1*a+k1*a+k1*2*a^2=

3*k1*a^2+2*k1*a=k1*(2*a+3*a^2) so k1=1/(2*a+3*a^2) and then

k2=-k1=-1/(2*a+3*a^2) and k3=-k1*(1+2*a)=-(1+2*a)/(2*a+3*a^2).

Then 1/(x^3+x^2+1)=

1/((x-a)*(2*a+3*a^2))-(x+1+2*a)/((x^2+a*x+x+a^2+a)*(2*a+3*a^2)).

Now you can do what you needed the partial fraction decomposition for, and just replace a^3=-1-a^2 until no a^3 or higher powers appear.

You can also write 1/(2*a+3*a^2)=k1+k2*a+k3*a^2 and multiply by (2*a+3*a^2) and after using a^3=-1-a^2 repeatedly, you get 1/(2*a+3*a^2)=-(3+11*a+2*a^2)/31 so you get

1/(x^3+x^2+1)=

-(3+11*a+2*a^2)/(31*(x-a))+(3+11*a+2*a^2)*(x+1+2*a)/(31*(x^2+a*x+x+a^2+a))

and the last term can have the numerator replaced by

(x*(2*a^2+11*a+3)+20*a^2+17*a-1).