New user? Sign up

Existing user? Log in

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, 2 months ago

$</code> ... <code>$</code>...<code>."> Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

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:$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Log in to reply

Thank You!

Most welcome!!

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

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

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

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 : $\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}$

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?

Click here for Newton's identities and click here too :).

@Nihar Mahajan – Hey, I am not an Idiot!

@Swapnil Das – Don't take it seriously. That was just for trolling :P :P

@Nihar Mahajan – I didn't take it seriously brother!

@Swapnil Das – It's quite funny how Nihar got to know about that site ;P

@A Brilliant Member – I knew that you would comment something here. :P

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

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

@Nihar Mahajan – Multitasker!

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

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

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

@Swapnil Das – Both are convenient for me. I just added this for variety of methods.

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

Correct.I was also going to say that.

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

Such questions r in class 9 RD Sharma book.

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestThe 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:$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$.

Log in to reply

Thank You!

Log in to reply

Most welcome!!

Log in to reply

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

Log in to reply

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

Log in to reply

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

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 : $\dfrac{\varphi_3}{\sigma_3} = \dfrac{3\sigma_3}{\sigma_3} = \boxed{3}$

Log in to reply

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?Log in to reply

Click here for Newton's identities and click here too :).

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

Log in to reply

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

Lol :P.Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

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

Log in to reply

Correct.I was also going to say that.

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

Log in to reply

Thank You!

Log in to reply

Such questions r in class 9 RD Sharma book.

Log in to reply