(A & sup2; - B & sup2;) + (3a-3b) how many factorizations are there~

(A & sup2; - B & sup2;) + (3a-3b) how many factorizations are there~

Physics search maniac,
（a2-b2)+(3a-3b)
=(a+b)(a-b)+3(a-b)
=(a-b)(a+b+3)

Factorization of 2 (a-b) & sup2; (a + b) - (B-A) & sup2; (3a-b)

The original formula 2 (a-b) & #178; (a + b) - (B-A) & #178; (3a-b)
= 2(a-b)²（a+b)-（a－b）²（3a－b）
=(a－b)²[2(a＋b)－(3a－b)]
=(a－b)²(3b－a)

Simplify 4A ^ 3 / (a ^ 4 + x ^ 4) + 1 / (a + x) + 2A / (a ^ 2 + x ^ 2) + 1 / (A-X)

4a^3/(a^4+x^4)+1/(a+x)+2a/(a^2+x^2)+1/(a-x) =4a^3/(a^4+x^4)+2a/(a^2+x^2)+[1/(a-x) )+1/(a+x）]=4a^3/(a^4+x^4)+[2a/(a^2+x^2)+2a/(a^2-x^2)]=4a^3/(a^4+x^4)+4a^3/(a^4-x^4)=8a^7/(a^8-x^8)

It is known that a = {2a-1,5-2a, - 3}, B = {a-1,4a + 1, a + 1}, a ∩ B = {- 3},

A ∩ B = {- 3}, so A-1 = - 3 or 4A + 1 = - 3. When A-1 = - 3, a = - 2; when 4A + 1 = - 3, a = - 1, a = - 2, a = {7,9, - 3} B = {- 3, - 7,5} a = - 1, a = {1,7, - 3} B = {- 2, - 3,2} so a = - 1 or - 2