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\(\quad If\quad a+b+c=0,\quad then\quad find\quad \frac { { a }^{ 2 } }{ bc } +\frac { { b }^{ 2 } }{ ca } +\frac { { c }^{ 2 } }{ ab }\)

Note by Swapnil Das
2 years, 1 month ago

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The same question was in my FIITJEE package. :p

Anyway, Here's the solution.

I cannot type in LaTeX, because I am operating from my iPad.

Taking the LCM of the 3 terms to be added, we get,

(abc)(a^3+b^3+c^3)/(abc)^2

=(a^3+b^3+c^3)./(abc)

Now, Since a+b+c=0, a^3+b^3+c^3=3abc.

=3abc/abc=3. QED.

Method 2

Jugaad method. Try 1+1-2 :P Mehul Arora · 2 years, 1 month ago

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@Mehul Arora :3 :3 :3 :3 :3 Cheers! Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan Cheers! I hope you like the 2nd Method. :P Mehul Arora · 2 years, 1 month ago

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The given expression can be converted to this,\[(a^3+b^3+c^3)/abc\].Now,because \(a+b+c=0,a^3+b^3+c^3=3abc\),hence answer is 3.Proof:\[a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)\]. Adarsh Kumar · 2 years, 1 month ago

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@Adarsh Kumar Thank You! Swapnil Das · 2 years, 1 month ago

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@Swapnil Das Most welcome!! Adarsh Kumar · 2 years, 1 month ago

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You need to specify that \(a,b,c \in \mathbb R - \{0\} \) otherwise it would be undefined. Krishna Sharma · 2 years, 1 month ago

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@Krishna Sharma Correct.I was also going to say that. Nihar Mahajan · 2 years, 1 month ago

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Let \(a+b+c= \sigma_1 \\ ab+bc+ac=\sigma_2 \\ abc = \sigma_3 \\ \varphi_3 = a^3+b^3+c^3 \)

By Newton's sums we have:

\[\varphi_3=\sigma_1^3-3\sigma_1\sigma_2 + 3\sigma_3 = (0)^3-3(0)\sigma_2+3\sigma_3 = 3\sigma_3\]

The given expression can be written as : \(\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}\) Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan And is the Newton's Sum method easier than the usual method? Swapnil Das · 2 years, 1 month ago

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@Swapnil Das It depends from person to person whether its easier or not. Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan I really wanna learn about Newton's Sum, well which method was easier to you? Swapnil Das · 2 years, 1 month ago

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@Swapnil Das Both are convenient for me. I just added this for variety of methods. Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan So Mr.Mathematician, @Nihar Mahajan , Thank U, but Can you tell me what are Newton's sum or provide a link to it so that I learn about them? Swapnil Das · 2 years, 1 month ago

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@Swapnil Das Click here for Newton's identities and click here too :). Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan Hey, I am not an Idiot! Swapnil Das · 2 years, 1 month ago

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@Swapnil Das It's quite funny how Nihar got to know about that site ;P Azhaghu Roopesh M · 2 years, 1 month ago

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@Azhaghu Roopesh M I knew that you would comment something here. :P Nihar Mahajan · 2 years, 1 month ago

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@Swapnil Das Don't take it seriously. That was just for trolling :P :P Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan I didn't take it seriously brother! Swapnil Das · 2 years, 1 month ago

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@Nihar Mahajan +1 for ur note & +10 for the other link. How did u get to that link?? Did anyone share the link with u or u just figured it out?? Aditya Kumar · 2 years, 1 month ago

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@Aditya Kumar hehe , actually once Pi Han Goh had mentioned that link in some note (I dont remember).Thats why I know about it. Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan Lol :P. @Pi Han Goh where did u get the link. How many hours do u stay online on brilliant @Nihar Mahajan ?? Aditya Kumar · 2 years, 1 month ago

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@Aditya Kumar I may "appear" to be attentive on brilliant , but behind the screen , I am actually doing some other work :) Nihar Mahajan · 2 years, 1 month ago

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@Nihar Mahajan Multitasker! Aditya Kumar · 2 years, 1 month ago

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Such questions r in class 9 RD Sharma book. Amit Kumar · 2 years, 1 month ago

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If a+b+c=0 Then,a^3+b^3+c^3-3abc=0 Therefore,a^3+b^3+c^3=3abc

Now in the question ,LCM of denominators will be "abc''

Thus ,on the numerator we get,a^3+b^3+c^3

Which is equal to 3abc

Therefore substituting it,we get the answer as 3...

Answer::3 Naitik Sanghavi · 2 years, 1 month ago

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@Naitik Sanghavi Thank You! Swapnil Das · 2 years, 1 month ago

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