Math, asked by rk2349349, 9 hours ago

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

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Answers

Answered by MrImpeccable
80

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!!!

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