Question Number 115508 by bemath last updated on 26/Sep/20
$${If}\begin{vmatrix}{{a}\:\:\:\:{a}^{\mathrm{2}} \:\:\:\:\:\:\:\mathrm{1}+{a}^{\mathrm{3}} }\\{{b}\:\:\:\:{b}^{\mathrm{2}} \:\:\:\:\:\:\:\mathrm{1}+{b}^{\mathrm{3}} }\\{{c}\:\:\:\:{c}^{\mathrm{2}} \:\:\:\:\:\:\:\mathrm{1}+{c}^{\mathrm{3}} }\end{vmatrix}=\:\mathrm{0} \\ $$$${a}\neq{b}\neq{c}\:\rightarrow\begin{cases}{{a}\:=?}\\{{b}=?\:}\\{{c}=?}\end{cases} \\ $$
Answered by bobhans last updated on 26/Sep/20
$$\Rightarrow\begin{vmatrix}{{a}\:\:\:\:{a}^{\mathrm{2}} \:\:\:\:\mathrm{1}}\\{{b}\:\:\:\:{b}^{\mathrm{2}} \:\:\:\:\mathrm{1}}\\{{c}\:\:\:\:{c}^{\mathrm{2}} \:\:\:\:\mathrm{1}}\end{vmatrix}+\:\begin{vmatrix}{{a}\:\:\:{a}^{\mathrm{2}} \:\:\:\:{a}^{\mathrm{3}} }\\{{b}\:\:\:\:{b}^{\mathrm{2}} \:\:\:{b}^{\mathrm{3}} \:}\\{{c}\:\:\:\:{c}^{\mathrm{2}} \:\:\:{c}^{\mathrm{3}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(−\mathrm{1}\right)\begin{vmatrix}{\mathrm{1}\:\:\:{a}^{\mathrm{2}} \:\:\:\:{a}}\\{\mathrm{1}\:\:\:{b}^{\mathrm{2}} \:\:\:\:{b}}\\{\mathrm{1}\:\:\:{c}^{\mathrm{2}} \:\:\:{c}}\end{vmatrix}+\:{abc}\begin{vmatrix}{\mathrm{1}\:\:\:{a}\:\:\:{a}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:{b}\:\:\:\:{b}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:{c}\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(−\mathrm{1}\right)^{\mathrm{2}} \begin{vmatrix}{\mathrm{1}\:\:\:\:{a}\:\:\:\:{a}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{b}\:\:\:\:\:{b}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{c}\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}+\:{abc}\:\begin{vmatrix}{\mathrm{1}\:\:\:\:{a}\:\:\:\:{a}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{b}\:\:\:\:\:{b}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{c}\:\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{abc}\right)\:\begin{vmatrix}{\mathrm{1}\:\:\:\:{a}\:\:\:\:{a}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{b}\:\:\:\:{b}^{\mathrm{2}} }\\{\mathrm{1}\:\:\:\:{c}\:\:\:\:{c}^{\mathrm{2}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{abc}\right)\:\begin{vmatrix}{\mathrm{1}\:\:\:\:\:\:\:\:\:\:\:{a}\:\:\:\:\:\:\:\:\:\:\:\:\:{a}^{\mathrm{2}} }\\{\mathrm{0}\:\:\:\:\:\:\:\:\:\:{b}−{a}\:\:\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\\{\mathrm{0}\:\:\:\:\:\:\:\:\:{c}−{a}\:\:\:\:\:{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{abc}\right)\:\begin{vmatrix}{{b}−{a}\:\:\:\:\:\:\:\:{b}^{\mathrm{2}} −{a}^{\mathrm{2}} }\\{{c}−{a}\:\:\:\:\:\:\:{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\end{vmatrix}=\:\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{abc}\right)\left[\left({b}−{a}\right)\left({c}−{a}\right)\left({c}+{a}\right)−\left({c}−{a}\right)\left({b}−{a}\right)\left({b}+{a}\right)\right]=\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}+{abc}\right)\left({b}−{a}\right)\left({c}−{a}\right)\:\left[\:{c}+{a}−{b}−{a}\right]\:=\:\mathrm{0} \\ $$$$\left(\mathrm{1}+{abc}\right)\left({b}−{a}\right)\left({c}−{a}\right)\left({c}−{b}\right)\:=\:\mathrm{0}\: \\ $$$${since}\:{a}\neq{b}\neq{c}\:;\:{so}\:{we}\:{get}\:{abc}\:=\:−\mathrm{1}\: \\ $$$${let}\:{a}\:=\:{k}\:,\:{b}\:=\:{l}\:\rightarrow\:{c}\:=\:−\frac{\mathrm{1}}{{k}.{l}} \\ $$$$ \\ $$
Commented by bemath last updated on 26/Sep/20
$${gave}\:{kudos} \\ $$
Answered by TANMAY PANACEA last updated on 26/Sep/20
$$\begin{vmatrix}{{a}}&{{a}^{\mathrm{2}} }&{\mathrm{1}}\\{{b}}&{{b}^{\mathrm{2}} }&{\mathrm{1}}\\{{c}}&{{c}^{\mathrm{2}} }&{\mathrm{1}}\end{vmatrix}+\begin{vmatrix}{{a}}&{{a}^{\mathrm{2}} }&{{a}^{\mathrm{3}} }\\{{b}}&{{b}^{\mathrm{2}} }&{{b}^{\mathrm{3}} }\\{{c}}&{{c}^{\mathrm{2}} }&{{c}^{\mathrm{3}} }\end{vmatrix}=\mathrm{0} \\ $$$$\begin{vmatrix}{{a}}&{{a}^{\mathrm{2}} }&{\mathrm{1}}\\{{b}}&{{b}^{\mathrm{2}} }&{\mathrm{1}}\\{{c}}&{{c}^{\mathrm{2}} }&{\mathrm{1}}\end{vmatrix}+{abc}\begin{vmatrix}{\mathrm{1}}&{{a}}&{{a}^{\mathrm{2}} }\\{\mathrm{1}}&{{b}}&{{b}^{\mathrm{2}} }\\{\mathrm{1}}&{{c}}&{{c}^{\mathrm{2}} }\end{vmatrix}=\mathrm{0} \\ $$$$\begin{vmatrix}{\mathrm{1}}&{{a}}&{{a}^{\mathrm{2}} }\\{\mathrm{1}}&{{b}}&{{b}^{\mathrm{2}} }\\{\mathrm{1}}&{{c}}&{{c}^{\mathrm{2}} }\end{vmatrix}+{abc}\begin{vmatrix}{\mathrm{1}}&{{a}}&{{a}^{\mathrm{2}} }\\{\mathrm{1}}&{{b}}&{{b}^{\mathrm{2}} }\\{\mathrm{1}}&{{c}}&{{c}^{\mathrm{2}} }\end{vmatrix}=\mathrm{0} \\ $$$${D}+{abc}×{D}=\mathrm{0} \\ $$$${D}\left(\mathrm{1}+{abc}\right)=\mathrm{0} \\ $$$${D}\neq\mathrm{0}\:\:\:{so}\:{abc}=−\mathrm{1} \\ $$$${let}\:{a}={k}\:\:\:{b}=\frac{\mathrm{1}}{{k}}\:\:\:{c}=−\mathrm{1} \\ $$$${i}\:{have}\:{just}\:{tried}… \\ $$
Commented by bemath last updated on 26/Sep/20
$${gave}\:{kudos}\:{sir}. \\ $$