Question

The answer of my last question is 2/5. You can take -2xy as -xy - xy then put value in (x-y)^2. Then u all can solve
Now another question.
Please don't give irrevelant answer, if u know then give answer with process or else leave.
Thanks a lot.​

Answer 1

Consider LHS.

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}}$

Since $\sf{\left(\dfrac{1}{a}\right)^m=\dfrac{1}{a^m},}$

$\begin{aligned}\longrightarrow\ \ &\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}}\\\\=\ \ &\sf{\dfrac{1}{(x^{a-b})^{\frac{1}{a-c}}}\cdot\dfrac{1}{(x^{b-c})^{\frac{1}{b-a}}}\cdot\dfrac{1}{(x^{c-a})^{\frac{1}{c-b}}}}\end{aligned}$

Since $\sf{\left(a^m\right)^n=a^{mn},}$

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=\dfrac{1}{x^{\frac{a-b}{a-c}}}\cdot\dfrac{1}{x^{\frac{b-c}{b-a}}}\cdot\dfrac{1}{x^{\frac{c-a}{c-b}}}}$

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=\dfrac{1}{x^{\frac{a-b}{a-c}}\cdot x^{\frac{b-c}{b-a}}\cdot x^{\frac{c-a}{c-b}}}}$

Since $\sf{a^m\cdot a^n=a^{m+n},}$

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=\dfrac{1}{x^{\frac{a-b}{a-c}+\frac{b-c}{b-a}+\frac{c-a}{c-b}}}}$

Let us simplify power of $\sf{x}$ in the denominator.

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=\dfrac{(a-b)(b-a)+(b-c)(c-b)+(c-a)(a-c)}{(a-c)(b-a)(c-b)}}$

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=\dfrac{-(a-b)^2-(b-c)^2-(c-a)^2}{-(a-b)(b-c)(c-a)}}$

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=\dfrac{(a-b)^2+(b-c)^2+(c-a)^2}{(a-b)(b-c)(c-a)}}$

$\begin{aligned}\longrightarrow\ \ &\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}}\\\\=\ \ &\sf{\dfrac{(a-b+b-c+c-a)^2-2((a-b)(b-c)+(b-c)(c-a)+(c-a)(a-b))}{(a-b)(b-c)(c-a)}}\end{aligned}$

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=-2\left(\dfrac{1}{a-b}+\dfrac{1}{b-c}+\dfrac{1}{c-a}\right)}$

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=-2\left(\dfrac{a-b+b-c+c-a}{(a-b)(b-c)(c-a)}\right)}$

$\longrightarrow\sf{\dfrac{a-b}{a-c}+\dfrac{b-c}{b-a}+\dfrac{c-a}{c-b}=0}$

Therefore,

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=\dfrac{1}{x^{0}}}$

$\longrightarrow\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=\dfrac{1}{1}}$

$\longrightarrow\underline{\underline{\sf{\left(\dfrac{1}{x^{a-b}}\right)^{\frac{1}{a-c}}\left(\dfrac{1}{x^{b-c}}\right)^{\frac{1}{b-a}}\left(\dfrac{1}{x^{c-a}}\right)^{\frac{1}{c-b}}=1}}}$

Hence Proved!