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

(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).