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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 · 2 years, 1 month ago

:3 :3 :3 :3 :3 Cheers! · 2 years, 1 month ago

Cheers! I hope you like the 2nd Method. :P · 2 years, 1 month ago

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:$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$. · 2 years, 1 month ago

Thank You! · 2 years, 1 month ago

Most welcome!! · 2 years, 1 month ago

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

Correct.I was also going to say that. · 2 years, 1 month ago

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 : $$\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}$$ · 2 years, 1 month ago

And is the Newton's Sum method easier than the usual method? · 2 years, 1 month ago

It depends from person to person whether its easier or not. · 2 years, 1 month ago

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

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

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? · 2 years, 1 month ago

Hey, I am not an Idiot! · 2 years, 1 month ago

It's quite funny how Nihar got to know about that site ;P · 2 years, 1 month ago

I knew that you would comment something here. :P · 2 years, 1 month ago

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

I didn't take it seriously brother! · 2 years, 1 month ago

+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?? · 2 years, 1 month ago

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

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

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

Multitasker! · 2 years, 1 month ago

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

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 · 2 years, 1 month ago