If a + b + c = 3, a2 + b2 + c2 = 6 and 1 + 1 + 1 = 1,

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If a + b + c = 3, a² + b² + c² = 6 and 1 + 1 + 1 = 1, where a, b, c are all non-zero, then ‘abc’ is equal to
a b c

a + b + c = 3; a² + b² + c² = 6
∴ (a + b + c)² = a² + b² + c² + 2ab + 2bc + 2ca
⇒ 3² = 6 + 2 (ab + bc + ca)
⇒ 9 – 6 = 2(ab + bc + ca)

⇒ ab + bc + ca =	3
	2

∴	1	+	1	+	1	= 1
	a		b		c

⇒	bc + ac + ab	= 1
	abc

⇒ abc = ab + bc + ca =	3
	2

If a + b + c = 3, a² + b² + c² = 6 and	1	+	1	+	1	= 1, where a, b, c are all non-zero, then ‘abc’ is equal to
	a		b		c