Prove that (x^a / x^b)^1/ab (x^b / x^c)^1/bc (x^c / x^a)^1/ca = 1

Exponential formulas of multiplication and division can be used to simplify such expressions.

Answer: Value of the L.H.S. comes out to be 1 thus the identity holds true.

Go through the proof and understand the solvation of this expression.

Explanation:

L.H.S = (x^a / x^b)^1/ab (x^b / x^c)^1/bc (x^c / x^a)^1/ca

On simplifying the L.H.S, we get:

(x^a / x^b)^1/ab (x^b / x^c)^1/bc (x^c / x^a)^1/ca

⇒ (x^{a - b})^1/ab (x^{b - c})^1/bc (x^{c - a})^1/ca 

⇒ (x^{(a - b)/ab}) (x^{(b - c)/bc}) (x^{(c - a)/ca})

⇒ (x^{[(a - b)/ab] + [(b - c)/bc] + [(c - a)/ca]})

⇒ x ^{(ac - bc + ab - ca + bc - ab) / abc}

⇒ x⁰ = 1

= R.H.S

Since, R.H.S = L.H.S, hence proved.