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Question Number 190407 by cortano12 last updated on 02/Apr/23

Answered by som(math1967) last updated on 02/Apr/23

$$\frac{b+c}{a}+1=\frac{c+a}{b}+1=\frac{a+b}{c}+1$$$$\Rightarrow \frac{a+b+c}{a}=\frac{a+b+c}{b}=\frac{a+b+c}{c}$$$$\Rightarrow \frac{1}{a}=\frac{1}{b}=\frac{1}{c}[\because (a+b+c)\ne 0]$$$$\Rightarrow a=b=c$$$$now\frac{a^3+b^3+c^3+6abc}{{(b+c)}^3}$$$$=\frac{a^3+a^3+a^3+6a^3}{{\left(2a\right)}^3}[\because a=b=c]$$$$=\frac{9a^3}{8a^3}=\frac{9}{8}ans$$

Commented by cortano12 last updated on 02/Apr/23

$$\frac{9}{8}\mathrm{sir}$$

Commented by som(math1967) last updated on 02/Apr/23

$$\underbrace{⋐}$$

Answered by horsebrand11 last updated on 02/Apr/23

$$let\{\begin{array}{l}b+c=pa\\ c+a=pb\\ a+b=pc\end{array}\Rightarrow 2(a+b+c)=p(a+b+c)$$$$p=2\Rightarrow \{\begin{array}{l}b+c=2a\\ c+a=2b\\ a+b=2c\end{array}\Rightarrow a=b=c=k$$$$\Rightarrow \frac{a^3+b^3+c^3+6abc}{{(b+c)}^3}=\frac{9k^3}{8k^3}=\frac{9}{8}$$

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