Question

If $\frac{loga}{b-c}=\frac{logb}{c-a}=\frac{logc}{a-b}$, then $a^{b+c}.b^{c+a}.c^{a+b}$=

Answer 1

$\frac{\log a}{b-c}=\frac{\log b}{c-a}=\frac{\log c}{a-b}=\lambda$ (let)

$\log a=\lambda (b-c),\log b=\lambda (c-a),\log c=\lambda (a-b)$

$a=e^{\lambda (b-c)},b=e^{\lambda (c-a)},c=e^{\lambda (a-b)}$

$a^{b+c}.b^{c+a}.c^{a+b}$

put the value of

$a,b,c$

$e^{\lambda (b-c)(b+c)\lambda (c+a)(c-a)\lambda (a+b)(a-b)}.e.e$

$e^{\lambda (b^2-c^2}.e^{\lambda (c^2-a^2)}.e^{\lambda (a^2-b^2)}$

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

$=e^o=1$