FIND THE ANSWER OF THIS QUESTION...............

It's really a nice problem. So we have 2 equations and 3 variables??? $a+b+c=0$ $a^2+b^2+c^2=\sqrt{74}$ We know that: $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ac)$ $0=\sqrt{74}+2(ab+bc+ac)$ $ab+bc+ac=-\dfrac{\sqrt{74}}{2}$ Let us assume a equation with $a,b,c$ as its roots so sum of roots is $0$ and sum of product of roots taken two at a time is $-\frac{\sqrt{74}}{2}$ .

Hence the equation is $x^3-\dfrac{\sqrt{74}}{2}x+k=0$ Here k is any constant. Putting $a$ in the equation: $a^3=\dfrac{\sqrt{74}}{2}a-k$ Similarly for $b$ and $c$ . The expression we have to find is $a^4+b^4+c^4$ $=a^3a+b^3b+c^3c$ $=a^2\frac{\sqrt{74}}{2}-ak+b^2\frac{\sqrt{74}}{2}-bk+c^2\frac{\sqrt{74}}{2}-ck$ $=\dfrac{\sqrt{74}}{2}(a^2+b^2+c^2)-k(a+b+c)$ $=\dfrac{74}{2}-0$ $=37$

(a+b+c)^2 = 0

--> (a^2) + (b^2) + (c^2) + 2*(ab + ac + bc) = 0; (a^2) + (b^2) + (c^2) = (74)^0.5

--> ab + ac +bc = -(1/2)*(74)^0.5

(ab + ac + bc)^2 = 74/4

--> (ab)^2 + 2ab(ac + bc) + (ac + bc)^2 = 37/2

--> (ab)^2 + (ac)^2 + (bc)^2 + 2[ (a^2)bc + a(b^2)c + ab(c^2) ] = 37/2

--> (ab)^2 + (ac)^2 + (bc)^2 + 2(abc)[ a + b + c ] = 37/2

By observation, 2(abc)[ a + b + c ]= 0 because a + b + c = 0

Hence, (ab)^2 + (ac)^2 + (bc)^2 = 37/2 = (ab + ac + bc)^2

a^4 + b^4 + c^4 is identical to: [[(a^2) + (b^2) + (c^2)]^2] -2[(ab)^2 + (ac)^2 + (bc)^2]

Thus, a^4 + b^4 + c^4= [(74^0.5)^2] - 2(37/2) = 74 - 37 = 37 // (QED)

this shud hav been a 80 point problem ..... also i saw another prob which was marked 100 pts ..... but it was worth only 60 point

Peace be upon you,

as a^4 + b^4 + c^4 = ( a^2 + b^2 + c^2)^2 - 2[(ab)^2 + (bc)^2 + (ac)^2],(1st),

as (a+b+c)^2 = 2(ab+bc+ac) + a^2 + b^2 + c^2

(0)^2 = 2(ab+bc+ac) + (74)^0.5

-(74)^0.5 = 2(ab+bc+ac)------> by squaring both sides,

you will get 74 = 4(ab)^2 + 4(bc)^2 + 4(ac)^2 + 8abc(a+b+c), as a+b+c =0,

37 = 2[(ab)^2 + (bc)^2 + (ac)^2],(2nd)------> substitute (2nd) into (1st),

a^4 +b^4 + c^4 = [(74)^0.5]^2 - 37 = 74 - 37 = 37....

thanks....