Question

ans fast

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Answer 1

Answer:

your question isn't proper.....

may be your question is.....

[(x)^a-b]^a+b × [(x)^b-a]^b+a × [(x)^c-a]^c+a = 1

(x)^a^2-b^2 × (x)^b^2-a^2 × (x)^c^2-a^2 = 1

we know that in multiplication when the bases are equal then there exponents are added....

(x)a^2-b^2+b^2-a^2+c^2-a^2 = 1

(x)^0 = 1

1 = 1

LHS = RHS

hope it helps you