Question

Prove that :(2a/2b)a+b x (2b/2c)b+c x (2c/2a)c+a =1 .

Answer 1

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Answer 2

Answer and Explanation:

To prove : $(\frac{2^a}{2^b})^{a+b}\times(\frac{2^b}{2^c})^{b+c}\times(\frac{2^c}{2^a})^{c+a}=1$

Proof :

Taking LHS,

$(\frac{2^a}{2^b})^{a+b}\times(\frac{2^b}{2^c})^{b+c}\times(\frac{2^c}{2^a})^{c+a}$

Apply $\frac{x^a}{x^b}=x^{a-b}$

$=(2^{a-b})^{a+b}\times(2^{b-c})^{b+c}\times(2^{c-a})^{c+a}$

Apply $(x^a)^b=x^{ab}$

$=2^{(a-b)(a+b)}\times2^{(b-c)(b+c)}\times2^{(c-a)(c+a)}$

We know, $(a-b)(a+b)=a^2-b^2$

$=2^{a^2-b^2}\times2^{b^2-c^2}\times2^{c^2-a^2}$

Apply $x^a\times x^b=x^{a+b}$

$=2^{a^2-b^2+b^2-c^2+c^2-a^2}$

$=2^{0}$

$=1$

= RHS

Hence proved.