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# if a+b+c=0 and a^2+b^2+c^2=1 then the value of a^4+b^4+c^4= ?

9 years ago

(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

9 years ago

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,