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Mathematics

Show that :

(ama−n)m−n×(ana−l)n−l×(ala−m)l−m\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} = 1

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Answer

Solving L.H.S. of the above equation :

⇒(ama−n)m−n×(ana−l)n−l×(ala−m)l−m⇒(am−(−n))m−n×(an−(−l))n−l×(al−(−m))l−m⇒(a(m+n))m−n×(a(n+l))n−l×(a(l+m))l−m⇒a(m+n)(m−n)×a(n+l)(n−l)×a(l+m)(l−m)⇒am2−n2×an2−l2×al2−m2⇒am2−n2+n2−l2+l2−m2⇒am2−m2−n2+n2−l2+l2⇒a0⇒1.\Rightarrow \Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} \\[1em] \Rightarrow (a^{m - (-n)})^{m - n} \times (a^{n - (-l)})^{n - l} \times (a^{l - (-m)})^{l - m} \\[1em] \Rightarrow (a^{(m + n)})^{m - n} \times (a^{(n + l)})^{n - l} \times (a^{(l + m)})^{l - m} \\[1em] \Rightarrow a^{(m + n)(m - n)} \times a^{(n + l)(n - l)} \times a^{(l + m)(l - m)} \\[1em] \Rightarrow a^{m^2 - n^2} \times a^{n^2 - l^2} \times a^{l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - n^2 + n^2 - l^2 +l^2 - m^2} \\[1em] \Rightarrow a^{m^2 - m^2 - n^2 + n^2 - l^2 + l^2} \\[1em] \Rightarrow a^0 \\[1em] \Rightarrow 1.

Since, L.H.S. = R.H.S. = 1.

Hence, proved that (ama−n)m−n×(ana−l)n−l×(ala−m)l−m\Big(\dfrac{a^m}{a^{-n}}\Big)^{m - n} \times \Big(\dfrac{a^n}{a^{-l}}\Big)^{n - l} \times (\dfrac{a^l}{a^{-m}}\Big)^{l - m} = 1.

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