Question

p.t (2a/2b)^a+b×(2b/2c)^b+c×(2c/2a)^c+a=1

Answer 1

Hope the ques is 

$((\frac{ 2^{a}}{ 2^{b}}) ^{a+b} ) * ((\frac{ 2^{b}}{ 2^{c}})^{b+c} ) * (( \frac{ 2^{c}}{ 2^{a}}) ^{c-a} ) = 1$
__________________________________________________________

Taking LHS

$2(^{a-b})( ^{a+b}) * 2(^{b-c})( ^{b+c}) * 2(^{c-a})( ^{c+a})$

$2^{a^{2}-b^{2} } * 2^{b^{2}-c^{2} } * 2^{c^{2}-a^{2} }$

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

$2^{0}$

1 = RHS
_______________________________________________________

☺☺☺ Hope this Helps ☺☺☺

Answer 2

Answer:

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.

Step-by-step explanation: