Does (6(t^2 -1))/(2(t^2 +1))=3?

Answer 1

[ 6 (t-1) (t+1) ] / [ 2 ( t-i)(t+i) ] is a complex number using i = sqrt(-1)

Answer 2

(6(t^2 -1))/(2(t^2 +1))=3
(6(t^2 -1)) = 3(2(t^2 +1))
6(t^2-1) = 6(t^2+1)
t^2-1 = t^2 + 1
-1 = 1
There is no value of t which will work.