Neha Sethi ma'am (Meritnation Expert),
You provided the answer to the following below question which was based finding numerical value in the expression.


But, the answer you provided was not having any numerical answer. Actually before you, SHRUTI TYAGI ma'am had answered the same question, whose answer was "1 or "one" " ... BUT, I was unable to understand the solution given by SHRUTI ma'am. Please assist properly.  

(x^y)^z = x^yz
similarly
x^yx^z = x^(y+z)

we have a case of
(x^u)^v (x^y)^z (x^p)^q 
=x^{(uv + yz + pq)}
u = 1/(a-b)
v= 1/(a-c)
y =1/(b-a)
z=1/(b-c)
p=1/(c-a)
q=1/(c-b)
uv + yz + pq  =1/{(a-b)(a-c)} + 1/{(b-a)(b-c)} +1/{(c-a)(c-b)}
= 1/{(a-b)(a-c)} -1/{(a-b)(b-c)} +1/{(a-c)(b-c)}  # (b-a)=-(a-b), (c-a)=-(a-c) (c-b) = -(b-c)
LCM of denominators above fractions (a-b)(a-c) , (b-c)(a-c), (a-c)(b-c) is (a-b)(b-c)(a-c)
1/{(a-b)(a-c)} =(b-c)/{(a-b)(b-c)(a-c)}
1/{(a-b)(b-c)} =(a-c)/{(a-b)(b-c)(a-c)}
1/{(a-c)(b-c)} =(a-b)/{(a-b)(b-c)(a-c)}
1/{(a-b)(a-c)} -1/{(a-b)(b-c)} +1/{(a-c)(b-c)} 
=(b-c)/{(a-b)(b-c)(a-c)}  - (a-c)/{(a-b)(b-c)(a-c)} + (a-b)/{(a-b)(b-c)(a-c)}
= { (b-c) -(a-c) + (a-b)}/{(a-b)(b-c)(a-c)}
=( b - c -a +c +a -b )/{(a-b)(b-c)(a-c)}
=(b -b -c +c -a +a)/{(a-b)(b-c)(a-c)}
=0/{(a-b)(b-c)(a-c)}
=0
uv + yz + pq = 0
Thus
x^{(uv + yz + pq)} = x⁰ = 1