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Question Number 188508 by Rupesh123 last updated on 02/Mar/23

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

$$a^3+b^3+c^3-3abc=0$$$$\Rightarrow (a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$$$$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$$$$[(a+b+c)\ne 0]$$$$\Rightarrow 2a^2+2b^2+2c^2-2ab-2bc-2ca=0$$$$\Rightarrow {(a-b)}^2+{(b-c)}^2+{(c-a)}^2=0$$$$forrealvalueofa,b,c$$$${(a-b)}^2=0\Rightarrow a-b=0\Rightarrow a=b$$$${(b-c)}^2\Rightarrow b=c$$$$\therefore a=b=c$$$$\frac{a^2+b^2+c^2}{{(a+b+c)}^2}=\frac{3a^2}{9a^2}=\frac{1}{3}$$

Commented by Rupesh123 last updated on 02/Mar/23

Excellent

Answered by Rasheed.Sindhi last updated on 02/Mar/23

$$a^3+b^3+c^3=3abc,a+b+c\ne 0$$$$\frac{a^2+b^2+c^2}{{(a+b+c)}^2}=?$$$$a^3+b^3+c^3-3abc=0$$$$(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0$$$$\Rightarrow a^2+b^2+c^2-ab-bc-ca=0$$$$a^2+b^2+c^2=ab+bc+ca$$$$▸\frac{a^2+b^2+c^2}{{(a+b+c)}^2}=\frac{a^2+b^2+c^2}{a^2+b^2+c^2+2(ab+bc+ca)}$$$$=\frac{a^2+b^2+c^2}{a^2+b^2+c^2+2(a^2+b^2+c^2)}$$$$=\frac{a^2+b^2+c^2}{3(a^2+b^2+c^2)}=\frac{1}{3}$$

Commented by Rupesh123 last updated on 02/Mar/23

Excellent

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