Decomposition factor: (AX by) ^ 3 + (by CZ) ^ 3 + (CZ ax) ^ 3

Decomposition factor: (AX by) ^ 3 + (by CZ) ^ 3 + (CZ ax) ^ 3

(ax-by)^3+(by-cz)^3+(cz-ax)^3
=((ax-by)+(by-cz))((ax-by)^2-(ax-by)(by-cz)+(by-cz)^2)+(cz-ax)^3
=(ax-cz)((ax-by)^2-(ax-by)(by-cz)+(by-cz)^2-(cz-ax)^2)
=(ax-cz)((ax-by)(ax-by-by+cz)+(by-cz+cz-ax)(by-cz-cz+ax))
=(ax-cz)((ax-by)(ax-2by+cz)+(by-ax)(by-2cz+ax))
=(ax-cz)(ax-by)(ax-2by+cz-by+2cz-ax)
=3(ax-cz)(ax-by)(cz-by)

Ax = by = CZ = 1, find (1 / 1 + A4) + (1 / 1 + B4) + (1 / 1 + C4) + (1 / 1 + x4) + (1 / 1 + Y4) + (1 / 1 + Z4) = (4:4 power)

When AX = by = CZ = 1, find the value of 1 / (1 + A ^ 4) + 1 / (1 + B ^ 4) + 1 / (1 + C ^ 4) + 1 / (1 + x ^ 4) + 1 / (1 + y ^ 4) + 1 / (1 + Z ^ 4)
really?
a=1/x,b=1/y,c=1/z
So 1 / (1 + A ^ 4) + 1 / (1 + x ^ 4) = 1 / (1 + 1 / x ^ 4) + 1 / (1 + x ^ 4) = x ^ 4 / (1 + x ^ 4) + 1 / (1 + x ^ 4) = 1
The other four items can also be worked out in pairs
So in the end
three

Factorization factor: (AX by) to the third power + (by CZ) to the third power - (AX CZ) to the third power

（ax-by)^3+(by-cz)^3-(ax-cz)^3=(ax-by+by-cz)[(ax-by)²-(ax-by)(by-cz)+(by-cz)²]-(ax-cz)³=(ax-cz)[(ax-by)²-(ax-by)(by-cz)+(by-cz)²-(ax-cz)²]=(ax-cz){(ax-by)²-(ax-by)(b...

Given x ^ 2 + y ^ 2 + Z ^ 2 = 1. A ^ 2 + B ^ 2 + C ^ 2 = 1... Verification: ax + by + CZ < = 1