Question

prove that : (2power a/2powerb)powera+b × (2powerb/2powerc)powerb+c × (2powerc/2powera)powerc + a = 1​

Answer 1

ANSWER:

To Prove::

$\:\:\bullet\:\:\left(\dfrac{2^a}{2^b}\right)^{a+b}\times\left(\dfrac{2^b}{2^c}\right)^{b+c}\times\left(\dfrac{2^c}{2^a}\right)^{c+a}=1$

Proof:

We need to prove that,

$\implies\left(\dfrac{2^a}{2^b}\right)^{a+b}\times\left(\dfrac{2^b}{2^c}\right)^{b+c}\times\left(\dfrac{2^c}{2^a}\right)^{c+a}=1$

Taking LHS,

$\implies\left(\dfrac{2^a}{2^b}\right)^{a+b}\times\left(\dfrac{2^b}{2^c}\right)^{b+c}\times\left(\dfrac{2^c}{2^a}\right)^{c+a}$

We know that,

$\hookrightarrow \dfrac{m^x}{m^y}=m^{x-y}$

So,

$\implies\left(\dfrac{2^a}{2^b}\right)^{a+b}\times\left(\dfrac{2^b}{2^c}\right)^{b+c}\times\left(\dfrac{2^c}{2^a}\right)^{c+a}$

$\implies\left(2^{a-b}\right)^{a+b}\times\left(2^{b-c}\right)^{b+c}\times\left(2^{c-a}\right)^{c+a}$

We know that,

$\hookrightarrow (m^x)^y=m^{xy}$

So,

$\implies\left(2^{a-b}\right)^{a+b}\times\left(2^{b-c}\right)^{b+c}\times\left(2^{c-a}\right)^{c+a}$

$\implies\left(2^{(a-b)(a+b)}\right)\times\left(2^{(b-c)(b+c)}\right)\times\left(2^{(c-a)(c+a)}\right)$

We know that,

$\hookrightarrow (x-y)(x+y)=x^2-y^2$

So,

$\implies\left(2^{(a-b)(a+b)}\right)\times\left(2^{(b-c)(b+c)}\right)\times\left(2^{(c-a)(c+a)}\right)$

$\implies\left(2^{a^2-b^2}\right)\times\left(2^{b^2-c^2}\right)\times\left(2^{c^2-a^2}\right)$

We also know that,

$\hookrightarrow m^x\times m^y=m^{x+y}$

So,

$\implies\left(2^{a^2-b^2}\right)\times\left(2^{b^2-c^2}\right)\times\left(2^{c^2-a^2}\right)$

$\implies2^{(a^2-b^2)+(b^2-c^2)+(c^2-a^2)}$

So,

$\implies2^{a^2\!\!\!/\:-b^2 \!\!\!/\: +b^2 \!\!\!/\: -c^2 \!\!\!/\: +c^2 \!\!\!/\: -a^2 \!\!\!/\: }$

$\implies2^{0}$

As, x^0=1. So,

$\implies2^0$

$\implies \bf1=RHS$

As, LHS=RHS,

HENCE PROVED!!!