Calculation 20083−2×20082−2006 20083+20082−2009.

Calculation 20083−2×20082−2006 20083+20082−2009.

Original formula =20082(2008−2)−2006
20082(2008+1)−2009=2006(20082−1)
2009(20082−1)=2006
2009.

The square of 2-2 to the fourth power of cube-2 to the 2008 power of...-2 to the 2009 power of 2=? There should be a decoupling process (preferably an additional explanation), The square of 2-2, the cubic of -2, the fourth power of -..., the 2008 power of -2, and the 2009 power of 2=? There should be a decoupling process (preferably an additional explanation), The square of 2-2, the cubic of -2, the fourth power of -......-2 to the 2009 power of 2=? There should be a decoupling process (preferably an additional explanation),

Because 2^2009=2*2^2008
So 2^2009-2^2008=(2-1)*2^2008=2^2008
Then 2^2008-2^2007=2^2007
.
2^3-2^2=2^2
So original formula =2+2^2=6
If the original equation =-2-2 square-2 cubic-2 fourth power -...-2 2008 power +2 2009 power =2