Question

100 Points ...
Solve as fast as u can...
But need a GR8 ANSWER..​

Answer 1

Answer:

if a+b+c=0 the (a+b+c)^2 = 0

a^2+b^2+c^2+2(ab+bc+ca) = 0

a^2+b^2+c^2 = -2(ab+bc+ca) squaring on both sides

a^4+b^4+c^4+2(a^2b^2+b^2c^2+c^2a^2) = 4(a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+ca^2b))

a^4+b^4+c^4 = 4(a^2b^2+b^2c^2+c^2a^2+2(ab^2c+bc^2a+ca^2b))-2(a^2b^2+b^2c^2+c^2a^2)

= 2[2(a^2b^2+b^2c^2+c^2a^2+(ab^2c+bc^2a+ca^2b))-(a^2b^2+b^2c^2+c^2a^2)]

= 2((a^2b^2+b^2c^2+c^2a^2+(ab^2c+bc^2a+ca^2b)))

Answer 2

hi mate,

solution,

a+b+c = 0

(a+b)² = (-c)²

a²+b²+2ab = c²

(a²+b²-c²)² = (-2ab)²

a⁴+b⁴+c⁴+2a²b²-2b²c²-2c²a² = 4a²b²

a⁴+b⁴+c⁴+2a²b²-4a²b²-2b²c²-2c²a² = 0

a⁴+b⁴+c⁴-2a²b²-2b²c²-2c²a²= 0

a⁴+b⁴+c⁴= 2a²b²+2b²c²+2c²a²

a⁴+b⁴+c⁴= 2(a²b²+b²c²+c²a²)

Substituting this value with

a⁴+b⁴+c⁴/a²b²+b²c²+c²a²

We get,

2(a²b²+b²c²+c²a²)/ a²b²+b²c²+c²a² = 2

i hope it helps you.