SOLUTION: a^2*(b^2+ab+a^2)*(b^2-ab+a^2)-(a^2+b^2)^3

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Question 1110935: a^2*(b^2+ab+a^2)*(b^2-ab+a^2)-(a^2+b^2)^3
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
ONE WAY:
A well known "special product" you are taught when learning about polynomials is
%28A-B%29%28A%2BB%29=A%5E2-B%5E2 .
A and B could be numbers, variables, polynomials or any mathematical expression.
We can use that to calculate that expression.
a%5E2%2A%28b%5E2%2Bab%2Ba%5E2%29%2A%28b%5E2-ab%2Ba%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22
%22=%22a%5E2%2A%28%28b%5E2%2Ba%5E2%29%5E2-a%5E2b%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22a%5E2%2A%28b%5E2%2Ba%5E2%29%5E2-a%5E4b%5E2-%28a%5E2%2Bb%5E2%29%28a%5E2%2Bb%5E2%29%5E2
%22=%22%28a%5E2-%28a%5E2%2Bb%5E2%29%29%2A%28b%5E2%2Ba%5E2%29%5E2-a%5E4b%5E2
%22=%22%28a%5E2-a%5E2-b%5E2%29%2A%28b%5E2%2Ba%5E2%29%5E2-a%5E4b%5E2
%22=%22-b%5E2%2A%28b%5E2%2Ba%5E2%29%5E2-a%5E4b%5E2
%22=%22-b%5E2%2A%28%28b%5E2%2Ba%5E2%29%5E2%2Ba%5E4%29
%22=%22-b%5E2%2A%28b%5E4%2Ba%5E4%2B2a%5E2b%5E2%2Ba%5E4%29
%22=%22highlight%28-b%5E2%2A%28b%5E4%2B2a%5E2b%5E2%2B2a%5E4%29%29

ANOTHER WAY:
From start to finish, we can write
a%5E2%2A%28b%5E2%2Bab%2Ba%5E2%29%2A%28b%5E2-ab%2Ba%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22
%22=%22a%5E2%2A%28%28a%5E3%29%5E2-%28b%5E3%29%5E2%29%2F%28a%5E2-b%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22a%5E2%2A%28a%5E6-b%5E6%29%2F%28a%5E2-b%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22a%5E2%2A%28a%5E4%2Ba%5E2b%5E2%2Bb%5E4%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22
%22=%22a%5E6%2Ba%5E4b%5E2%2Ba%5E2b%5E4-%28a%5E6%2B3a%5E4b%5E2%2B3a%5E2b%5E4%2Bb%5E3%29
%22=%22a%5E6%2Ba%5E4b%5E2%2Ba%5E2b%5E4-a%5E6-3a%5E4b%5E2-3a%5E2b%5E4-b%5E3
%22=%22highlight%28-2a%5E4b%5E2-2a%5E2b%5E4-b%5E3=-b%5E2%2A%28b%5E4%2B2a%5E2b%5E2%2B2a%5E4%29%29
How does it work?
Less used special polynomial products say that
A%5E3-B%5E3=%28A-B%29%28A%5E2%2BAB%2BB%5E2%29 and A%5E3%2BB%5E3=%28A%2BB%29%28A%5E2-AB%2BB%5E2%29 .
So, A%5E2%2BAB%2BB%5E2=%28A%5E3-B%5E3%29%2F%28A-B%29 and A%5E2-AB%2BB%5E2=%28A%5E3%2BB%5E3%29%2F%28A%2BB%29 .
As you learn about sequences and series,
you could realize that a%5E2%2Bab%2Bb%5E2 and a%5E2-ab%2Bb%5E2
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%5E2%2A%28b%5E2%2Bab%2Ba%5E2%29%2A%28b%5E2-ab%2Ba%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22
Now we can use %28A-B%29%28A%2BB%29=A%5E2-B%5E2 to continue with
%22=%22a%5E2%2A%28%28a%5E3%29%5E2-%28b%5E3%29%5E2%29%2F%28a%5E2-b%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
%22=%22a%5E2%2A%28a%5E6-b%5E6%29%2F%28a%5E2-b%5E2%29-%28a%5E2%2Bb%5E2%29%5E3
But now we see that a%5E6-b%5E6=%28a%5E2%29%5E3-%28b%5E2%29%5E3
is also a difference of cubes, just like A%5E3-B%5E3=%28A-B%29%28A%5E2%2BAB%2BB%5E2%29 ,
and if %28A%5E3-B%5E3%29%2F%28A-B%29=A%5E2%2BAB%2BB%5E2 , then