Question 1116335
A) You can use the"special product {{{(A-B)(A^2+AB+B^2)=A^3-B^3}}} twice,
with {{{system(A=x^3,B=y^3)}}} the first time, and {{{system(A=x,B=y)}}} the second time.
{{{x^9-y^9=(x^3-y^3)(x^6+x^3y^3+y^6)=highlight((x-y)(x^2+xy+y)(x^6+x^3y^3+y^6))}}}
The last expression is completely factorized.
A polynomial of the form {{{A^2+AB+B^2}}} is positive as long as {{{AB<>0}}} ,
so it cannot be factored.
 
B) {{{x^4-5x^3+5x^2+5x-6}}}
takes the value {{{0}}} for {{{x=1}}} and for {{{x=-1}}} ,
so {{{(x-1)}}} and {{{(x+1)}}} are factors.
Dividing by both, you find
{{{x^4-5x^3+5x^2+5x-6=highlight((x-1)(x+1)(x^2-5x+1))}}} 
The last expression is completely factorized,
unless you wanted to include {{{(x-(5+sqrt(21))/2)}}} and {{{(x-(5-sqrt(21))/2)}}} as factors,
because the zeros of {{{x^2-5x+1}}} are the irrational numbers {{{(5 +- sqrt(21))/2}}}