If a + b + c = 0, then prove the following(a) (b + c) (b − c) + a(a

1.	If a + b + c = 0, then prove the following(a) (b + c) (b − c) + a(a + 2b) = 0(b) a(a2 − bc) + b(b2 − ca) + c(c2 − ab) = 0(c) a(b2 + c2) + b(c2 + a2) + c(a2 + b2) = −3abc(d)
Answer» If a + b + c = 0, then prove the following (a) (b + c) (b − c) + a(a + 2b) = 0 (b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0 (c) a(b² + c²) + b(c² + a²) + c(a² + b²) = −3abc (d)

If a + b + c = 0, then prove the following(a) (b + c) (b − c) + a(a + 2b) = 0(b) a(a2 − bc) + b(b2 − ca) + c(c2 − ab) = 0(c) a(b2 + c2) + b(c2 + a2) + c(a2 + b2) = −3abc(d)

Answer»

If a + b + c = 0, then prove the following

(a) (b + c) (b − c) + a(a + 2b) = 0

(b) a(a² − bc) + b(b² − ca) + c(c² − ab) = 0

(d)

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If a + b + c = 0, then prove the following(a) (b + c) (b − c) + a(a + 2b) = 0(b) a(a2 − bc) + b(b2 − ca) + c(c2 − ab) = 0(c) a(b2 + c2) + b(c2 + a2) + c(a2 + b2) = −3abc(d)

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