Prove that x^4/((x-k)(x^2+k^2)) cannot be partially factioned. Thanks?

Prove that x^4/[(x－k)(x^2+k^2)] cannot be partially factioned. Thanks?
Sol
題目錯誤
設x^4/[(x－k)(x^2+k^2)]=x+a/(x－k)+(bx+c)/(x^2+k^2)
x^4=x(x－k)(x^2+k^2)+a(x^2+k^2)+(bx+c)(x－k)
當x=k
k^4=a(k^2+k^2)
a=k^2/2
x^4=x(x－k)(x^2+k^2)+k^2(x^2+k^2)/2+(bx+c)(x－k)
2x^4=2x(x－k)(x^2+k^2)+k^2(x^2+k^2)+2(bx+c)(x－k)
當x=ki
2k^4=2(bki+c)(ki－k)
k^3=(bki+c)(i－1)
k^3=－bk－bki+ci-c
bki－ci+k^3+bk+c=0
i(bk－c)+(k^3+bk+c)=0
bk－c=0．k^3+bk+c=0
(bk－c)+(k^3+bk+c)=0
2bk+k^3=0
b=－k^2/2
c=bk=－k^3/2
x^4/[(x－k)(x^2+k^2)]=x+(k^2/2)/(x－k)－[(k^2x+k^3)/2]/(x^2+k^2)

Let x⁴ / [(x - k)(x² + k²)] = a / (x - k) + (bx + c) / (x² + k²)

x⁴ = ax² + ak² + bx² - bkx + cx - ck = (a + b)x² +(-bk + c)x + ak² - ck

It is invalid

Q.E.D. #