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Question

If $a+b+c=6,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2},\mathrm{then}\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=__________.$

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Solution

$\mathrm{Given}:\phantom{\rule{0ex}{0ex}}\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2}...\left(1\right)\phantom{\rule{0ex}{0ex}}a+b+c=6...\left(2\right)\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\mathrm{Now},\phantom{\rule{0ex}{0ex}}\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\phantom{\rule{0ex}{0ex}}\mathrm{Adding}\mathrm{and}\mathrm{subtracting}3,\mathrm{we}\mathrm{get}\phantom{\rule{0ex}{0ex}}=\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+3-3\phantom{\rule{0ex}{0ex}}=\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+1+1+1-3\phantom{\rule{0ex}{0ex}}=\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+\frac{a}{a}+\frac{b}{b}+\frac{c}{c}-3\phantom{\rule{0ex}{0ex}}=\left(\frac{a}{a}+\frac{b}{a}+\frac{c}{a}\right)+\left(\frac{a}{b}+\frac{b}{b}+\frac{c}{b}\right)+\left(\frac{a}{c}+\frac{b}{c}+\frac{c}{c}\right)-3\phantom{\rule{0ex}{0ex}}=\left(\frac{a+b+c}{a}\right)+\left(\frac{a+b+c}{b}\right)+\left(\frac{a+b+c}{c}\right)-3\phantom{\rule{0ex}{0ex}}=\left(\frac{6}{a}\right)+\left(\frac{6}{b}\right)+\left(\frac{6}{c}\right)-3\left(\mathrm{From}\left(2\right)\right)\phantom{\rule{0ex}{0ex}}=\frac{6}{a}+\frac{6}{b}+\frac{6}{c}-3\phantom{\rule{0ex}{0ex}}=6\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3\phantom{\rule{0ex}{0ex}}=6\left(\frac{3}{2}\right)-3\left(\mathrm{From}\left(1\right)\right)\phantom{\rule{0ex}{0ex}}=9-3\phantom{\rule{0ex}{0ex}}=6$ Hence, if $a+b+c=6,\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{2},\mathrm{then}\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}=\overline{)6}.$

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