Question

(3^a/3^b)^a+b×(3^b/3^c)^b+c×(3^c/3^a)c+a​

Answer 1

Answer :-

1

Given to find the value of :-

$\bigg(\dfrac{3^a}{3^b} \bigg)^{a+b} \times\bigg(\dfrac{3^b}{3^c} \bigg)^{b+c} \times\bigg(\dfrac{3^c}{3^a} \bigg)^{c+a}$

Formulae implemented:-

Here the Laws of Exponents can be used they were:-

$\dfrac{a^m}{a^n} = a^{m-n}$

$a^m\times a^n \times a^p = a^{m+n+p}$

$(a^m)^n = a^{mn}$

$a^0 = 1$

and here the Algebraic identity also used that is ,

$(a+b)(a-b) =a^2-b^2$

By using these formulae we can solve the given question .

Solution :-

$\bigg(\dfrac{3^a}{3^b} \bigg)^{a+b} \times\bigg(\dfrac{3^b}{3^c} \bigg)^{b+c} \times\bigg(\dfrac{3^c}{3^a} \bigg)^{c+a}$

$\bigg(3^{a-b}\bigg)^{a+b}\times \bigg(3^{b-c}\bigg)^{b+c}\times\bigg(3^{c-a}\bigg)^{c+a} [\dfrac{a^m}{a^n} = a^{m-n}]$

$\bigg(3^{(a-b)(a+b)}\bigg)\times\bigg( 3^{(b+c)(b-c)}\bigg)\times\bigg( 3^{(c+a)(c-a)}\bigg) [(a^m)^n = a^{mn}]$

$\bigg(3^{a^2-b^2}\bigg)\times \bigg(3^{b^2-c^2}\bigg)\times \bigg(3^{c^2-a^2}\bigg) [(a+b)(a-b) = a^2-b^2]$

$\bigg(3^{a^2-b^2+b^2-c^2+c^2-a^2}\bigg) [ a^m\times a^n \times a^p = a^{m+n+p}]$

$\bigg(3^{\not{a^2}\not{-b^2}\not{+b^2}\not{-c^2}\not{+c^2}\not{-a^2}}\bigg)$

$3^0$

$1$

So, the value of ,

${\boxed{\bigg(\dfrac{3^a}{3^b} \bigg)^{a+b} \times\bigg(\dfrac{3^b}{3^c} \bigg)^{b+c} \times\bigg(\dfrac{3^c}{3^a} \bigg)^{c+a} = 1}}$

Note:-

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