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Algebra

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9 Jun 2008 12:50:50 IST

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343

Sum of powers

None

Given and find the value of

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pink_ele

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Joined: 16 Jan 2007

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9 Jun 2008 15:13:27 IST

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(a+b+c)=0
(a+b+c)^2=0
a^2+b^2+c^2= -2(ab+bc+ac)
1=4(ab+bc+ac)^2
1=4 {b^2c^2+a^2b^2+a^2c^2)+8abc(a+b+c)
1/2=2{b^2c^2+a^2b^2+a^2c^2)
now
(a^2+b^2+c^2)^2=a^4+b^4+c^4+2{b^2c^2+a^2b^2+a^2c^2)
a^4+b^4+c^4=1-1/2=1/2...............:)

SUNDEEP ALLAMRAJU

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9 Jun 2008 15:39:20 IST

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There was a power cut in my house.So,I didn't give the solution before.Anyway,Here's my method,

c=-(a+b),put this in a²+b²+c²=1,we get,2(a²+b²+ab)=1.

Now,a⁴+b⁴+(a+b)⁴=2(a⁴+b⁴)+4ab(a²+b²)+6a²b²

=2[(a²+b²)²-2a²b²+2ab(a²+b²)+3a²b²]=2[(a²+b²)+ab]²=2(1/2)²=1/2.

Hari Shankar

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9 Jun 2008 17:44:17 IST

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A bit less beefy:

$0 = (a^3+b^3+c^3)(a+b+c) = \sum a^4 + \sum ab (a^2+b^2)$

$= \sum a^4 + \sum ab(1-c^2) = \sum a^4 + \sum ab - abc\sum a$

$= \sum a^4 + \sum ab$

Hence $\sum a^4 = -\sum ab = \frac{1}{2}$

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