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(a + b +c)2 = 0
a2 + b2 + c2 +2(ab + bc + ca) = 0
2(ab + bc + ca) = -1
squaring both the sides
4(a2b2 + b2c2 + c2a2 + 2(ab2c + bc2a +ca2b ) = 1
4(a2b2 + b2c2 + c2a2 + 2abc (a + b + c ) = 1
since a+b+c = 0 , therefore 2abc(a+b+c) becomes 0
therefore a2b2 + b2c2 + c2a2 = 0.25 -----------(1)-
(a2 + b2 + c2)2 = a4 + b4 + c4 + 2( a2b2 + b2c2 + a2c2 ) ------------(2)-
substituting the value of (1)- in (2)-
therefore a4 + b4 + c4 becomes 0.5
Hi Rupali,
a2+b2+c2 = (a+b+c)2 - 2Σ(ab)
So 1 = 0 - 2Σ(ab).
Or Σ(ab) = -1/2.
Now a4+b4+c4 = (a2+b2+c2)2 - 2(a2b2+b2c2+c2a2).
= (a2+b2+c2)2 - 2[ (ab+bc+ca)2 - 2abc(a+b+c) ]
= 1 - 2[ (-1/2)2 - 0 ]
= 1 - 2[1/4] = 1/2.
You can also learn some identities in the above working.
Hope that helps.
Best Regards,
Ashwin (IIT Madras).
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