Question

20.If a + b + c = 0, then prove that a4 + b4 + c4 = 2(b2c2 + c2a2 + a2b2)​

Answer 1

answer

ok

step by step ok

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)