# Help me!

Please do help me with this question, Thanks!

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

Note by Swapnil Das
4 years, 8 months ago

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Such questions r in class 9 RD Sharma book.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

:3 :3 :3 :3 :3 Cheers!

- 4 years, 8 months ago

Cheers! I hope you like the 2nd Method. :P

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Correct.I was also going to say that.

- 4 years, 8 months 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 : $\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}$

- 4 years, 8 months ago

And is the Newton's Sum method easier than the usual method?

- 4 years, 8 months ago

It depends from person to person whether its easier or not.

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Both are convenient for me. I just added this for variety of methods.

- 4 years, 8 months 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?

- 4 years, 8 months ago

- 4 years, 8 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??

- 4 years, 8 months ago

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

- 4 years, 8 months ago

Lol :P. @Pi Han Goh where did u get the link. How many hours do u stay online on brilliant @Nihar Mahajan ??

- 4 years, 8 months ago

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

- 4 years, 8 months ago

- 4 years, 8 months ago

Hey, I am not an Idiot!

- 4 years, 8 months ago

It's quite funny how Nihar got to know about that site ;P

- 4 years, 8 months ago

I knew that you would comment something here. :P

- 4 years, 8 months ago

Don't take it seriously. That was just for trolling :P :P

- 4 years, 8 months ago

I didn't take it seriously brother!

- 4 years, 8 months 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...

- 4 years, 8 months ago

Thank You!

- 4 years, 8 months 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:$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

- 4 years, 8 months ago

Thank You!

- 4 years, 8 months ago

Most welcome!!

- 4 years, 8 months ago