Hello mates I have a question. Please solve it fast
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Hello mates I have a question. Please solve it fast
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Answer:
Taking Base X as common
X^[(a-b) a+b]*[(b-c) b+c]*[(c-a) c+a]
(a-b)(a+b)*(b-c)(b+c)*(c-a)(c+a)
=>(a²-b²)(b²-c²)(c²-a²)
{ since a+b* a-b= a²-b²}
since every variable's square is present 2 times : once positive and once negative.. so they cancel each other (example- +a²-a² = 0)
therefore ultimately
X^0=1
since any number to the power 0 is = 1
Answer:
L.H.S. = R.H.S. , Proved.
Step-by-step explanation:
Formula Used :
See attachment for detailed solution.