Question Number 198849 by sonukgindia last updated on 25/Oct/23
Commented by mr W last updated on 25/Oct/23
$${for}\:{a},{b},\:{c}\:\in{R}\:{no}\:{solution}! \\ $$
Answered by deleteduser1 last updated on 25/Oct/23
$${a}+{b}+{c}=\mathrm{30};\left({ab}+{bc}+{ca}\right)=\frac{{abc}}{\mathrm{4}};{abc}=\mathrm{1000} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{250} \\ $$$${a},{b},{c}\:{are}\:{roots}\:{of}\:{x}^{\mathrm{3}} −\mathrm{30}{x}^{\mathrm{2}} +\mathrm{250}{x}−\mathrm{1000}=\mathrm{0} \\ $$$$\Rightarrow\left({a},{b},{c}\right)=\left(\mathrm{20},\mathrm{5}+\mathrm{5}{i},\mathrm{5}−\mathrm{5}{i}\right)\:{upto}\:{permutations}. \\ $$
Answered by mr W last updated on 25/Oct/23
$$\frac{{a}+{b}+{c}}{\mathrm{3}}=\mathrm{10} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{30} \\ $$$$\sqrt[{\mathrm{3}}]{{abc}}=\mathrm{10} \\ $$$$\Rightarrow{abc}=\mathrm{1000} \\ $$$$\frac{\mathrm{3}}{\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{c}}}=\mathrm{12} \\ $$$$\frac{\mathrm{3}{abc}}{{ab}+{bc}+{ca}}=\mathrm{12} \\ $$$${ab}+{bc}+{ca}=\frac{\mathrm{3}{abc}}{\mathrm{12}}=\frac{\mathrm{3}×\mathrm{1000}}{\mathrm{12}}=\mathrm{250} \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{250} \\ $$$$ \\ $$$${a},{b},{c}\:{are}\:{roots}\:{of}\:{x}^{\mathrm{3}} −\mathrm{30}{x}^{\mathrm{2}} +\mathrm{250}{x}−\mathrm{1000}=\mathrm{0} \\ $$$$\left({x}−\mathrm{20}\right)\left({x}^{\mathrm{2}} −\mathrm{10}{x}+\mathrm{50}\right)=\mathrm{0} \\ $$$$\Rightarrow{x}=\mathrm{20},\:\mathrm{5}\pm\mathrm{5}{i} \\ $$$$\Rightarrow{a},{b},{c}=\left(\mathrm{20},\:\mathrm{5}+\mathrm{5}{i},\:\mathrm{5}−\mathrm{5}{i}\right) \\ $$