How to factorize x ^ n-y ^ n and x ^ n + y ^ n

How to factorize x ^ n-y ^ n and x ^ n + y ^ n

x^n-y^n
=(x-y)[x^(n-1)+x^(n-2)y+x^(n-3)y^2+...+x^2y(n-3)+xy^(n-2)+y^(n-1)]
Note: that is, the sum of times of each item in brackets is n - 1. Multiply X - y to verify directly
If it's a plus,
When n is odd, you can directly set X ^ n + y ^ n = x ^ n - (- y) ^ n,
When n is an even number, there is always x ^ n + y ^ n > 0, and there is no real root, so it cannot be decomposed in the real number field, such as x ^ 2 + y ^ 2

Using factorization method to explore the value of (1-1 / 2 & # 178;) (1-1 / 3 & # 178;) (1 - / 4 & # 178;)... (1-1 / N & # 178;)
Given X & # 178; + 2x + 1 + Y & # 178; - 8y + 16 = 0, find the value of Y / X. -Do this by the way

（1-1/2²）（1-1/3²）（1-/4²）...（1-1/n²）
=(1-1/2)(1+1/2)(1-1/3)(1+1/3)(1-1/4)(1+1/4) *.*(1-1/n)(1+1/n)
= 1/2*3/2*2/3*4/3*3/4*5/4*.*(n-1)/n*(n+1)/n
= 1/2*(n+1)/n
= (n+1)/(2n)
x²+2x+1+y²-8y+16=0
(x+1)^2+(y-4)^2=0
x=-1,y=4
y/x=4/(-1)=-4

Please factorize 2 (a-b) ^ 3 - (B-A) ^ 2
Please decompose 2 (a-b) ^ 3 - (B-A) ^ 2 and write out the process of solving the problem. Thank you!

The original formula = 2 (a-b) & sup3; - (a-b) & sup2;
=(a-b)²[2(a-b)-1]
=(a-b)²(2a-2b-1)