Question

plz solve this sum in five minutes it is very compulsory .Show that:-​

Answer 1

ANSWER:

To Prove::

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

Proof:

We need to prove that,

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

Taking LHS,

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

We know that,

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

So,

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

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

We know that,

$\hookrightarrow (m^z)^y=m^{zy}$

So,

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

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

We know that,

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

So,

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

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

We also know that,

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

So,

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

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

So,

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

$\implies x^{0}$

As, z^0=1. So,

$\implies x^0$

$\implies \bf1=RHS$

As, LHS=RHS,

HENCE PROVED!!!