IIT JEE , AIEEE Forum - Mathematics , Algebra

hsbhatt

Given and find the value of

pink_ele

(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...............:)

allamraju

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.

hsbhatt

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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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.

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.