If a, b, c are positive real numbers, show that:a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1a−1b×b−1c×c−1a=1
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Given,
a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1a−1b×b−1c×c−1a=1
Solving L.H.S :
⇒a−1b×b−1c×c−1a⇒(1a)b×(1b)c×(1c)a⇒(ba)×(cb)×(ac)⇒(ba)×(cb)×(ac)⇒1⇒1.\Rightarrow \sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{1}{a}\Big)b} \times \sqrt{\Big(\dfrac{1}{b}\Big)c} \times \sqrt{\Big(\dfrac{1}{c}\Big)a} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{b}{a}\Big)} \times \sqrt{\Big(\dfrac{c}{b}\Big)} \times \sqrt{\Big(\dfrac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{\Big(\dfrac{b}{a}\Big) \times \Big(\dfrac{c}{b}\Big) \times \Big(\dfrac{a}{c}\Big)} \\[1em] \Rightarrow \sqrt{1} \\[1em] \Rightarrow 1.⇒a−1b×b−1c×c−1a⇒(a1)b×(b1)c×(c1)a⇒(ab)×(bc)×(ca)⇒(ab)×(bc)×(ca)⇒1⇒1.
Hence proved, a−1b×b−1c×c−1a=1\sqrt{a^{-1}b} \times \sqrt{b^{-1}c} \times \sqrt{c^{-1}a} = 1a−1b×b−1c×c−1a=1.
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