Question

(m-a)(m-b)(m-c).......(m-z)=?

Answer 1

answer will be 0 as at a point the equation formed will be (m-m)
hope this helps

Answer 2

Notice the following patterns:

(m -a)(m-b) = [m^2 - ma - mb + (ab)] = [ m^2 - (a+b)m + (ab)]

[ m^2 - (a+b)m + (ab)] [m - c ]  = 

m^3 - (a+b)m^2 + (ab)m - cm^2 + (ac)m + (bc)m - abc =

m^3 - (a + b + c)m^2 + (ab + ac + bc)m - abc

Let (a,b,c,d,e.......z)  be denoted as SetZ.....so we have..... 

(m-a)(m-b)(m-c)......(m-z)  =

m^26 - (sum of all elements in SetZ taken one at a time)m^25 + (sum of all elements in SetZ taken two at a time)m^24 - (sum of all elements of SetZ taken three at a time)m^23 +.........+ (sum of all elements in SetZ taken 24 at a time)m^2 -(sum of all elements of SetZ taken 25 at a time)m + (the product of all elements in SetZ)

I believe this is correct......!!!