Question 1110935
ONE WAY:
A well known "special product" you are taught when learning about polynomials is
{{{(A-B)(A+B)=A^2-B^2}}} .
A and B could be numbers, variables, polynomials or any mathematical expression.
We can use that to calculate that expression.
{{{a^2*(b^2+ab+a^2)*(b^2-ab+a^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*((b^2+a^2)+ab)*((b^2+a^2)-ab)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*((b^2+a^2)^2-a^2b^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*(b^2+a^2)^2-a^4b^2-(a^2+b^2)(a^2+b^2)^2}}}
{{{"="}}}{{{(a^2-(a^2+b^2))*(b^2+a^2)^2-a^4b^2}}}
{{{"="}}}{{{(a^2-a^2-b^2)*(b^2+a^2)^2-a^4b^2}}}
{{{"="}}}{{{-b^2*(b^2+a^2)^2-a^4b^2}}}
{{{"="}}}{{{-b^2*((b^2+a^2)^2+a^4)}}}
{{{"="}}}{{{-b^2*(b^4+a^4+2a^2b^2+a^4)}}}
{{{"="}}}{{{highlight(-b^2*(b^4+2a^2b^2+2a^4))}}}
 
ANOTHER WAY:
From start to finish, we can write
{{{a^2*(b^2+ab+a^2)*(b^2-ab+a^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*((a^3-b^3)/(a-b))*((a^3+b^3)/(a+b))-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*((a^3)^2-(b^3)^2)/(a^2-b^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*(a^6-b^6)/(a^2-b^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*(a^4+a^2b^2+b^4)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^6+a^4b^2+a^2b^4-((a^2)^3+3(a^2)^2b^2+3a^2(b^2)^2+(b^2)^3)}}}
{{{"="}}}{{{a^6+a^4b^2+a^2b^4-(a^6+3a^4b^2+3a^2b^4+b^3)}}}
{{{"="}}}{{{a^6+a^4b^2+a^2b^4-a^6-3a^4b^2-3a^2b^4-b^3}}}
{{{"="}}}{{{highlight(-2a^4b^2-2a^2b^4-b^3=-b^2*(b^4+2a^2b^2+2a^4))}}}
How does it work?
Less used special polynomial products say that
{{{A^3-B^3=(A-B)(A^2+AB+B^2)}}} and {{{A^3+B^3=(A+B)(A^2-AB+B^2)}}} .
So, {{{A^2+AB+B^2=(A^3-B^3)/(A-B)}}} and {{{A^2-AB+B^2=(A^3+B^3)/(A+B)}}} .
As you learn about sequences and series,
you could realize that {{{a^2+ab+b^2}}} and {{{a^2-ab+b^2}}}
are the sums of the three first terms of geometric sequences,
and could prove the same equivalent expressions for those sums.
We can use those alternate expressions to help calculate.
{{{a^2*(b^2+ab+a^2)*(b^2-ab+a^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*((a^3-b^3)/(a-b))*((a^3+b^3)/(a+b))-(a^2+b^2)^3}}}
Now we can use {{{(A-B)(A+B)=A^2-B^2}}} to continue with
{{{"="}}}{{{a^2*((a^3)^2-(b^3)^2)/(a^2-b^2)-(a^2+b^2)^3}}}
{{{"="}}}{{{a^2*(a^6-b^6)/(a^2-b^2)-(a^2+b^2)^3}}}
But now we see that {{{a^6-b^6=(a^2)^3-(b^2)^3}}}
is also a difference of cubes, just like {{{A^3-B^3=(A-B)(A^2+AB+B^2)}}} ,
and if {{{(A^3-B^3)/(A-B)=A^2+AB+B^2}}} , then
{{{(a^6-b^6)/(a^2-b^2)=((a^2)^3-(b^2)^3)/(a^2-b^2)=(a^2)^2+(a^2)(b^2)+(b^2)^2=a^4+a^2b^2+b^4}}}