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
6 years ago
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Easy Math Editor
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The given expression can be converted to this,(a3+b3+c3)/abc.Now,because a+b+c=0,a3+b3+c3=3abc,hence answer is 3.Proof:a3+b3+c3−3abc=(a+b+c)(a2+b2+c2−ab−bc−ca).
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Thank You!
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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
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:3 :3 :3 :3 :3 Cheers!
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Cheers! I hope you like the 2nd Method. :P
Let a+b+c=σ1ab+bc+ac=σ2abc=σ3φ3=a3+b3+c3
By Newton's sums we have:
φ3=σ13−3σ1σ2+3σ3=(0)3−3(0)σ2+3σ3=3σ3
The given expression can be written as : σ3φ3=σ33σ3=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?
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Click here for Newton's identities and click here too :).
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Hey, I am not an Idiot!
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Don't take it seriously. That was just for trolling :P :P
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I didn't take it seriously brother!
It's quite funny how Nihar got to know about that site ;P
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I knew that you would comment something here. :P
+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??
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hehe , actually once Pi Han Goh had mentioned that link in some note (I dont remember).Thats why I know about it.
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Lol :P. @Pi Han Goh where did u get the link. How many hours do u stay online on brilliant @Nihar Mahajan ??
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I may "appear" to be attentive on brilliant , but behind the screen , I am actually doing some other work :)
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Multitasker!
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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I really wanna learn about Newton's Sum, well which method was easier to you?
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Both are convenient for me. I just added this for variety of methods.
You need to specify that a,b,c∈R−{0} otherwise it would be undefined.
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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
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Thank You!
Such questions r in class 9 RD Sharma book.