Question

a+b+c=0 then (a4+b4+c4)/(a2+b2+c2)2​

Answer 1

Answer:

Answer is 2

Step-by-step explanation:

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

Answer 2

Answer:

Squaring on both sides of the relation,

= (a² + b² + c²)² = [-2(bc + ca + ab)²]

= 4 {b²c² + c²a² + a²b² + 2} {bc. ca + ca. ab + ab. bc}

= 4 (b²c² + c²a² + a²b²) + 8abc (a + b + c)

= 4 (b²c² + c²a² + a²b²),

∵ a + b + c = 0

∴ 2 (b²c² + c²a² + a²b²) = $\frac{1}{2}$ (a² + b² + c²)²

And also (a² + b² + c²)² = $(a^{4} + b^{4} +c^{4} )$ + 2 (b²c² + c²a² + a²b²),

⇒ 4 (b²c² + c²a² + a²b²) = $(a^{4} + b^{4} +c^{4} )$ + 2 (b²c² + c²a² + a²b²)

Hence,  $(a^{4} + b^{4} +c^{4} )$ = 2 (b²c² + c²a² + a²b²)