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

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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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## Comments

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

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Thank You!

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Most welcome!!

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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

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

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

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

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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 :).

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@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

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

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

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

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Correct.I was also going to say that.

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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

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Thank You!

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Such questions r in class 9 RD Sharma book.

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