Question 1009757
{{{x^5y-xy}}}...we can write the term {{{x^5y}}} as {{{x^4*xy}}}  

so,
 {{{x^4*xy-xy}}} ....both terms have {{{xy}}} in common, so we can factor out {{{xy}}}

{{{xy(x^4-1)}}}.... write the term {{{x^4}}} as {{{(x^2)^2}}} and {{{1}}} as  {{{1^2}}}

{{{xy((x^2)^2-1^2)}}}....now we have difference of squares, factor it  

{{{xy((x^2-1)(x^2+1))}}}....one more time we have difference of squares, factor it 

{{{xy((x-1)(x+1)(x^2+1))}}}




{{{5a^3-10a^2+3a-6}}}....group first two terms together and second two terms together

{{{(5a^3-10a^2)+(3a-6)}}} ...from a group of first two terms factor out {{{5a^2}}}, from a second group factor out {{{3}}}

{{{5a^2(a-2)+3(a-2)}}} 

{{{(5a^2+3)(a-2)}}}