Question

if a+b+c =3,
${a}^{2} + {b + {c}^{2} } = 6$
and 1/a+1/b+1/c =1 where a,b,c are non zero then abc is

plz give the answer fastly​

Answer 1

$Given \: a+b+c = 3 \: ---(1)$

$a^{2} + b^{2} + c^{2} = 6 \: ---(2)$

$2ab+2bc+2ca = (a+b+c)^{2} - (a^{2}+b^{2}+c^{2})$

$\implies 2(ab+bc+ca) = 3^{2} - 6$

$\implies ab+bc+ca = \frac{9-6}{2} \\= \frac{3}{2}\: --(3)$

$Now , \frac{1}{a} + \frac{1}{b}+ \frac{1}{c} = 1 \:(given )$

$\implies \frac{bc+ac+ab}{abc} = 1$

$\implies bc+ac+ab = abc$

$\implies \frac{3}{2} = abc\: [From \:(3) ]$

Therefore.,

$\red { Value \:of \:abc} \green {=\frac{3}{2}}$

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