Question

Show that ^ is = 1
rational numbers​

Answer 1

SOLUTION:

$\text{We need to prove that,}\\\\:\longrightarrow\dfrac{[x^{(a+b)}]^2\times[x^{(b+c)}]^2\times[x^{(c+a)}]^2}{(x^ax^bx^c)^4}=1\\\\\text{Solving LHS,}\\\\:\implies\dfrac{[x^{(a+b)}]^2\times[x^{(b+c)}]^2\times[x^{(c+a)}]^2}{(x^ax^bx^c)^4}\\\\\text{We know that,}\\\\:\hookrightarrow p^m\times p^n=p^{m+n}\\\\:\implies\dfrac{[x^{(a+b)}]^2\times[x^{(b+c)}]^2\times[x^{(c+a)}]^2}{(x^{(a+b+c)})^4}\\\\\text{We know that,}\\\\:\hookrightarrow p^{m^n}=p^{mn}\\\\:\implies\dfrac{[x^{2(a+b)}]\times[x^{2(b+c)}]\times[x^{2(c+a)}]}{(x^{4(a+b+c)})}\\\\:\implies\dfrac{[x^{2(a+b)+2(b+c)+2(c+a)}]}{(x^{4(a+b+c)})}\\\\:\implies\dfrac{[x^{2(a+b+b+c+c+a)}]}{x^{4(a+b+c)}}\\\\:\implies\dfrac{x^{2(2a+2b+2c)}}{x^{4(a+b+c)}}\\\\:\implies\dfrac{x^{4(a+b+c)}}{x^{4(a+b+c)}}\\\\\bf{:\implies 1=RHS}\\\\\text{\bf{Hence Proved!!!}}$

Formulae Used:

• $p^m\times p^n=p^{m+n}$
• $p^{m^n}=p^{mn}$