Please do help me with this question, **Thanks!**

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

Please do help me with this question, **Thanks!**

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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestThe 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 · 1 year, 11 months ago

Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

:3 :3 :3 :3 :3 Cheers!Log in to reply

– Mehul Arora · 1 year, 11 months ago

Cheers! I hope you like the 2nd Method. :PLog in to reply

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 · 1 year, 11 months ago

Log in to reply

– Swapnil Das · 1 year, 11 months ago

Thank You!Log in to reply

– Adarsh Kumar · 1 year, 11 months ago

Most welcome!!Log in to reply

You need to specify that \(a,b,c \in \mathbb R - \{0\} \) otherwise it would be undefined. – Krishna Sharma · 1 year, 11 months ago

Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

Correct.I was also going to say that.Log in to reply

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 · 1 year, 11 months ago

Log in to reply

– Swapnil Das · 1 year, 11 months ago

And is the Newton's Sum method easier than the usual method?Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

It depends from person to person whether its easier or not.Log in to reply

– Swapnil Das · 1 year, 11 months ago

I really wanna learn about Newton's Sum, well which method was easier to you?Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

Both are convenient for me. I just added this for variety of methods.Log in to reply

@Nihar Mahajan ,

So Mr.Mathematician,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 · 1 year, 11 months agoLog in to reply

Click here for Newton's identities and click here too :). – Nihar Mahajan · 1 year, 11 months ago

Log in to reply

– Swapnil Das · 1 year, 11 months ago

Hey, I am not an Idiot!Log in to reply

– Azhaghu Roopesh M · 1 year, 11 months ago

It's quite funny how Nihar got to know about that site ;PLog in to reply

– Nihar Mahajan · 1 year, 11 months ago

I knew that you would comment something here. :PLog in to reply

– Nihar Mahajan · 1 year, 11 months ago

Don't take it seriously. That was just for trolling :P :PLog in to reply

– Swapnil Das · 1 year, 11 months ago

I didn't take it seriously brother!Log in to reply

– Aditya Kumar · 1 year, 11 months 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??Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

hehe , actually once Pi Han Goh had mentioned that link in some note (I dont remember).Thats why I know about it.Log in to reply

@Pi Han Goh where did u get the link. How many hours do u stay online on brilliant @Nihar Mahajan ?? – Aditya Kumar · 1 year, 11 months ago

Lol :P.Log in to reply

– Nihar Mahajan · 1 year, 11 months ago

I may "appear" to be attentive on brilliant , but behind the screen , I am actually doing some other work :)Log in to reply

– Aditya Kumar · 1 year, 11 months ago

Multitasker!Log in to reply

Such questions r in class 9 RD Sharma book. – Amit Kumar · 1 year, 11 months ago

Log in to reply

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 · 1 year, 11 months ago

Log in to reply

– Swapnil Das · 1 year, 11 months ago

Thank You!Log in to reply