Question

If $7 \log _{a b c}\left(a^3+b^3+c^3\right)=3 k\left(\frac{1+\log _3(a b c)}{\log _3(a b c)}\right)$ and $(a b c)^{a+b+c}=1$ and $k=\frac{m}{n}$ in lowest form, where $m, n \in N$ then find the value $(m+n)$.

Answer 1

$ a+b+c=0 $
$\Rightarrow a ^3+ b ^3+ c ^3=3 abc $
$\text { L.H.S. }=7 \log _{a b c}(3 a b c) $
$\text { R.H.S. }=3 k \frac{\log _3(3 a b c)}{\log _3(a b c)} $
$=3 k \log _{ abc }(3 abc ) $
So $7 \log _{a b c}(3 a b c)=3 k \log _{a b c}(3 a b c) $
$ \therefore k=\frac{7}{3} \equiv \frac{m}{n}$
$\therefore( m + n )=7+3=10$