(9+1)(9的二次方+1)(9的四次方+1)=K分之9的八次方-1 則K=?

(9-1)(9+1)(9^2+1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^2-1)(9^2+1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^4-1)(9^4+1)=((9^8-1)/k)(9-1) ==>(9^8-1)/8=(9^8 -1)/k ==>k=8