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Algebra

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8 Jul 2008 19:12:06 IST

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Calling 9th and 10th graders

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$\text{If} \ x+y+z = 0, \text{Prove that} \\ \\<br/>(x^2+xy+y^2)^3+(y^2+yz+z^2)^3+(z^2+zx+x^2)^3 = 3 (x^2+xy+y^2)(y^2+yz+z^2)(z^2+zx+x^2)$

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Comments (14)

Chris21

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8 Jul 2008 19:59:36 IST

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its of the form

so prove the RHS of the formula as zero

Hari Shankar

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8 Jul 2008 20:01:27 IST

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pls type out the whole solution

Shreya

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8 Jul 2008 20:21:06 IST

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iam not a ninth or tenth grader, can i answer?

Shreya

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8 Jul 2008 20:37:18 IST

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plz tell me , can i answer?

Rahul Duggal

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9 Jul 2008 16:34:15 IST

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sir can you tell me how to prove

®µD®A

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9 Jul 2008 17:19:06 IST

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sorry

Hari Shankar

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9 Jul 2008 18:01:21 IST

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you have made it undefined my multiplying and dividing by zero!

®µD®A

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9 Jul 2008 18:03:22 IST

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sorry

Hari Shankar

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9 Jul 2008 18:13:42 IST

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I agree that that is the identity and not

$x^2+y^2+z^2 - xy-yz-zx = \frac{x^3+y^3+z^2-3xyz}{x+y+z}$

as it does not hold when x+y+z = 0

®µD®A

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9 Jul 2008 18:14:41 IST

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OK. Thank you sir

Hari Shankar

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9 Jul 2008 18:15:31 IST

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nice Latexing, by the way. I hope others follow your example.

Hari Shankar

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9 Jul 2008 18:17:02 IST

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and rahul, if you are unable to prove that 2(x^2+y^2+z^2)+xy+yz+zx = 0 , maybe you have to try some other approach

Hari Shankar

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10 Jul 2008 11:26:48 IST

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shall I give the go-ahead to the seniors to jump in?

abhishek sinha

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10 Jul 2008 14:08:05 IST

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entering into pic : ( with due consent )

let , $b=y^2+yz +z^2 \and\$ and

now we form ( a- b ) which has clearly a ( x + y + z) factor hence zero .as

a-b = as ( x+y +z =0 )

so a= b

similarly we can prove that b= c

So LHS = 3a^3 = RHS ( proved )

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