Question Number 190407 by cortano12 last updated on 02/Apr/23
Answered by som(math1967) last updated on 02/Apr/23
![((b+c)/a)+1=((c+a)/b)+1=((a+b)/c)+1 ⇒((a+b+c)/a)=((a+b+c)/b)=((a+b+c)/c) ⇒(1/a)=(1/b)=(1/c) [∵(a+b+c)≠0] ⇒a=b=c now ((a^3 +b^3 +c^3 +6abc)/((b+c)^3 )) =((a^3 +a^3 +a^3 +6a^3 )/((2a)^3 )) [∵a=b=c] =((9a^3 )/(8a^3 ))=(9/8)ans](https://www.tinkutara.com/question/Q190410.png)
$$\frac{{b}+{c}}{{a}}+\mathrm{1}=\frac{{c}+{a}}{{b}}+\mathrm{1}=\frac{{a}+{b}}{{c}}+\mathrm{1} \\ $$$$\Rightarrow\frac{{a}+{b}+{c}}{{a}}=\frac{{a}+{b}+{c}}{{b}}=\frac{{a}+{b}+{c}}{{c}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{a}}=\frac{\mathrm{1}}{{b}}=\frac{\mathrm{1}}{{c}}\:\:\left[\because\left({a}+{b}+{c}\right)\neq\mathrm{0}\right] \\ $$$$\Rightarrow{a}={b}={c} \\ $$$${now}\:\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{6}{abc}}{\left({b}+{c}\right)^{\mathrm{3}} } \\ $$$$=\frac{{a}^{\mathrm{3}} +{a}^{\mathrm{3}} +{a}^{\mathrm{3}} +\mathrm{6}{a}^{\mathrm{3}} }{\left(\mathrm{2}{a}\right)^{\mathrm{3}} }\:\left[\because{a}={b}={c}\right] \\ $$$$=\frac{\mathrm{9}{a}^{\mathrm{3}} }{\mathrm{8}{a}^{\mathrm{3}} }=\frac{\mathrm{9}}{\mathrm{8}}{ans} \\ $$
Commented by cortano12 last updated on 02/Apr/23
$$\:\frac{\mathrm{9}}{\mathrm{8}}\:\mathrm{sir} \\ $$
Commented by som(math1967) last updated on 02/Apr/23
$$\:\underbrace{\Subset} \\ $$
Answered by horsebrand11 last updated on 02/Apr/23
$$\:{let}\:\begin{cases}{{b}+{c}={pa}}\\{{c}+{a}={pb}}\\{{a}+{b}={pc}}\end{cases}\:\Rightarrow\mathrm{2}\left({a}+{b}+{c}\right)={p}\left({a}+{b}+{c}\right) \\ $$$$\:{p}=\mathrm{2}\:\Rightarrow\begin{cases}{{b}+{c}=\mathrm{2}{a}}\\{{c}+{a}=\mathrm{2}{b}}\\{{a}+{b}=\mathrm{2}{c}}\end{cases}\:\Rightarrow{a}={b}={c}\:={k} \\ $$$$\:\Rightarrow\frac{{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{6}{abc}}{\left({b}+{c}\right)^{\mathrm{3}} }\:=\:\frac{\mathrm{9}{k}^{\mathrm{3}} }{\mathrm{8}{k}^{\mathrm{3}} }\:=\frac{\mathrm{9}}{\mathrm{8}} \\ $$