on the grounds which you have not rather shown what you pick in brackets, one might desire to assume you propose: ((a hundred twenty five-2*5) Mod 3^3) / (3^2 -(-3^2)) ((a hundred twenty five-10) Mod 27) / (9 - 9) (one hundred fifteen Mod 27) / 0 7 / 0 it is considered as infinity in some circles. TexMav
Answers & Comments
Verified answer
the denominator is part of the
'difference of squares' equation
(9-5)=(3+sqrt5)(3-sqrt5)
multiply by
(3+sqrt5) / (3+sqrt5)
(2+sqrt3)(3+sqrt5) / (9-5)
6+(2sqrt5)(3sqrt3)+5 / 4
(6+6sqrt15+5)/4
(11+6sqrt15) / 4
on the grounds which you have not rather shown what you pick in brackets, one might desire to assume you propose: ((a hundred twenty five-2*5) Mod 3^3) / (3^2 -(-3^2)) ((a hundred twenty five-10) Mod 27) / (9 - 9) (one hundred fifteen Mod 27) / 0 7 / 0 it is considered as infinity in some circles. TexMav