Question

I am not able to proceed after this step
please help!

Answer 1

Well after

a(m - n) = d(n^2 -n - m^2 + m)

a(m - n) - d(n^2 - m^2 - n + m) = 0

Take - common

a(m - n) + d(m^2 - n^2 -m + n) = 0

a (m - n) + d { (m+n)(m - n)- (m - n)} = 0

a (m - n) + d(m - n){( m + n) - 1} = 0

Now take ( m - n ) common from whole

( m - n) { a + d( m + n ) - 1} = 0

a + ( m + n)d - 1 = 0 / ( m - n )

a + ( m + n)d - 1 = 0

Hence. A(m + n) = 0 is proved



Thanks