Question

Hello mates I have a question. Please solve it fast​

Answer 1

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 2

Answer:

L.H.S. = R.H.S. , Proved.

Step-by-step explanation:

Formula Used :

$x^{a^{b}} = x^{a.b} \\\\(a+b)(a-b)=a^{2} -b^{2} \\\\x^{a}.x^{b}= x^{a+b}\\\\a^{0}= 1$

See attachment for detailed solution.