Question

show that (x-y)3 + (y-z)3 + (z-x) 3 = 3 (x-y) (y-z) (z-x)

Answer 1

heya..........NAMSKAR

#It depends on how much you know already. It is quite easy if you know about the connection between factors and roots. First, notice that (without doing the complete expansion) you can see that the cubed terms cancel;

# for example, the x^3 from the first term cancels the -x^3 from the third term, etc. So, the left-hand-side is a quadratic in x, y and z, in the sense their highest powers are 2.

# Look at the left-hand-side L as a function of x; it is quadratic in x and vanishes when x = y or x = z [because when x = y it is 0^3 + (x-z)^3 + (z-x)^3 = 0, using x instead of y in the second term].

# So, the quadratic L factors as L = b*(x-y)(x-z), where b does not depend on x. Note that we can write this as L = c*(x-y)*(z-x), where c = -b; the c-form is closest to what we want, so let's use it.

#Similarly, as a function of y, L vanishes when y = x or y = z, so (y-x) and (y-z) are both factors of L; that is, our previous c must have y-z as a factor.


#Finally, we get L = d*(x-y)*(y-z)*(z-x), where d is a constant. Now we can determine d by substituting some convenient values such as z=0, y=1, z=2 into both sides. 





tysm...@kundan