Show that the equation 1/x-a +1/x-b +1/x-c =0 can have a pair of equal roots if a=b=c

The numerator becomes: (x-b)(x-c) + (x-a)(x-c) + (x-b)(x-c) 

We don't care about the bottom because if a fraction = 0, it's the numerator that must = 0 (ie, you can't divide by any number to make it 0) 
x^2 - cx - bx + bc + x^2 - ax - cx + ac + x^2 - bx - cx + bc = 0 
3x^2 - 2cx - 2bx - 2ax + ac + bc + ab= 0 
Quad formula 
x = (2c + 2b + 2a) +/- sqrt((2a+2b+2c)^2 - (4)(3)(ac+bc+ab)) / 6 
x = (2a + 2b + 2c) +/- 2sqrt(a^2 - ab -ac + b^2 - bc + b^2) / 6 

For double roots 

a^2 - ab -ac + b^2 - bc + c^2 = 0 
a^2 + b^2 + c^2 - (ab + bc + ac) = 0 
This only works if ab = a^2, bc = b^2 and ac = c^2... which means a = b = c