Question

if a+b+c=6 and 1/a+1/b+1/c=3/2, then find a/b+b/a+b/c+c/b+a/c+c/a

Answer 1

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Answer 2

Answer:

$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=6$

Step-by-step explanation:

$Given ,\\ a+b+c=6---(1)\\\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}--(2)$

$Multiply \: equations \: (1)\: and (2)\\we \: get$

$(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=6\times \frac{3}{2}$

$\implies a(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\\+b(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\\+c(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=9$

$\implies \frac{a}{a}+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{b}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{c}{c}=9$

$\implies 1+\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+1+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1=9$

$\implies \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3=9$

$\implies \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=9-3$

$Therefore,\\ \frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=6$

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