Question Number 49298 by peter frank last updated on 05/Dec/18
Answered by Kunal12588 last updated on 05/Dec/18
$$\left.{b}\right)\:\mathrm{3}{ac}\left({a}+{c}\right)+{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \\ $$$$=\mathrm{3}{ac}\left({a}+{c}\right)+\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{abc} \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{ac}\left({b}\right)+\mathrm{3}{ac}\left({a}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{ac}\left({a}+{b}+{c}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}+\mathrm{3}{ac}\right) \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}+\mathrm{2}{ca}\right) \\ $$
Commented by Kunal12588 last updated on 05/Dec/18
$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)+\mathrm{3}{abc} \\ $$
Commented by Kunal12588 last updated on 05/Dec/18
$${special}\:{case} \\ $$$${when}\:{a}+{b}+{c}=\mathrm{0}\: \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{3}{abc} \\ $$$${just}\:{reminded}\:{when}\:{I}\:{first}\:{seen}\:{this}\:{identity} \\ $$
Answered by Kunal12588 last updated on 05/Dec/18
$$\left({a}\right)\:\mathrm{27}{a}^{\mathrm{3}} {b}^{\mathrm{6}} −\mathrm{8}{c}^{\mathrm{3}} \\ $$$$=\left(\mathrm{3}{ab}^{\mathrm{2}} \right)^{\mathrm{3}} −\left(\mathrm{2}{c}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{3}{ab}^{\mathrm{2}} −\mathrm{2}{c}\right)\left(\mathrm{9}{a}^{\mathrm{2}} {b}^{\mathrm{4}} +\mathrm{6}{ab}^{\mathrm{2}} +\mathrm{4}{c}^{\mathrm{2}} \right) \\ $$$${as}\:{a}^{\mathrm{3}} −{b}^{\mathrm{3}} =\left({a}−{b}\right)\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right) \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 05/Dec/18
$$\left.{a}\right)\left(\mathrm{3}{ab}^{\mathrm{2}} \right)^{\mathrm{3}} −\left(\mathrm{2}{c}\right)^{\mathrm{3}} \\ $$$$=\left(\mathrm{3}{ab}^{\mathrm{2}} −\mathrm{2}{c}\right)\left\{\left(\mathrm{3}{ab}^{\mathrm{2}} \right)^{\mathrm{2}} +\mathrm{3}{ab}^{\mathrm{2}} ×\mathrm{2}{c}+\left(\mathrm{2}{c}\right)^{\mathrm{2}} \right\} \\ $$$$=\left(\mathrm{3}{ab}^{\mathrm{2}} −\mathrm{2}{c}\right)\left(\mathrm{9}{a}^{\mathrm{2}} {b}^{\mathrm{4}} +\mathrm{6}{ab}^{\mathrm{2}} {c}+\mathrm{4}{c}^{\mathrm{2}} \right) \\ $$$$\left.{b}\right){b}^{\mathrm{3}} +\left({a}+{c}\right)^{\mathrm{3}} \\ $$$$=\left({b}+{a}+{c}\right)\left\{{b}^{\mathrm{2}} −{b}\left({a}+{c}\right)+\left({a}+{c}\right)^{\mathrm{2}} \right\} \\ $$$$=\left({a}+{b}+{c}\right)\left({b}^{\mathrm{2}} −{ab}−{bc}+{a}^{\mathrm{2}} +\mathrm{2}{ac}+{c}^{\mathrm{2}} \right) \\ $$$$=\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}+\mathrm{2}{ac}\right) \\ $$$$\left.{c}\right){a}^{\mathrm{2}} \left({b}^{\mathrm{3}} −{c}^{\mathrm{3}} \right)+{b}^{\mathrm{2}} \left({c}^{\mathrm{3}} −{a}^{\mathrm{3}} \right)+{c}^{\mathrm{2}} \left({a}^{\mathrm{3}} −{b}^{\mathrm{3}} \right) \\ $$$$={a}^{\mathrm{2}} \left({b}−{c}\right)\left({b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} \right)+{b}^{\mathrm{2}} {c}^{\mathrm{3}} −{b}^{\mathrm{3}} {c}^{\mathrm{2}} +{a}^{\mathrm{3}} {c}^{\mathrm{2}} −{a}^{\mathrm{3}} {b}^{\mathrm{2}} \\ $$$$={a}^{\mathrm{2}} \left({b}−{c}\right)\left({b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} {c}^{\mathrm{2}} \left({b}−{c}\right)−{a}^{\mathrm{3}} \left({b}^{\mathrm{2}} −{c}^{\mathrm{2}} \right) \\ $$$$={a}^{\mathrm{2}} \left({b}−{c}\right)\left({b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} \right)−{b}^{\mathrm{2}} {c}^{\mathrm{2}} \left({b}−{c}\right)−{a}^{\mathrm{3}} \left({b}+{c}\right)\left({b}−{c}\right) \\ $$$$=\left({b}−{c}\right)\left({a}^{\mathrm{2}} {b}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}+{a}^{\mathrm{2}} {c}^{\mathrm{2}} −{b}^{\mathrm{2}} {c}^{\mathrm{2}} −{a}^{\mathrm{3}} {b}−{a}^{\mathrm{3}} {c}\right) \\ $$$$=\left({b}−{c}\right)\left\{−{b}^{\mathrm{2}} \left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)+{a}^{\mathrm{2}} {c}\left({c}−{a}\right)+{a}^{\mathrm{2}} {b}\left({c}−{a}\right)\right\} \\ $$$$=\left({b}−{c}\right)\left\{\left({c}−{a}\right)\left(−{b}^{\mathrm{2}} {c}−{b}^{\mathrm{2}} {a}\right)+\left({c}−{a}\right)\left({a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}\right)\right\} \\ $$$$=\left({b}−{c}\right)\left({c}−{a}\right)\left\{−{b}^{\mathrm{2}} {c}−{b}^{\mathrm{2}} {a}+{a}^{\mathrm{2}} {c}+{a}^{\mathrm{2}} {b}\right\} \\ $$$$=\left({b}−{c}\right)\left({c}−{a}\right)\left\{{c}\left({a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right)+{ab}\left({a}−{b}\right)\right\} \\ $$$$=\left({b}−{c}\right)\left({c}−{a}\right)\left\{\left({a}−{b}\right)\left({ac}+{bc}+{ab}\right)\right\} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\left({ab}+{bc}+{ac}\right) \\ $$
Commented by peter frank last updated on 06/Dec/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}. \\ $$
Answered by $@ty@m last updated on 06/Dec/18
$${Alternative}\:{way}\:{for}\:\left({c}\right) \\ $$$${The}\:{given}\:{identity}: \\ $$$$\begin{vmatrix}{{a}^{\mathrm{2}} }&{{b}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\\{{a}^{\mathrm{3}} }&{{b}^{\mathrm{3}} }&{{c}^{\mathrm{3}} }\\{\mathrm{1}}&{\mathrm{1}}&{\mathrm{1}}\end{vmatrix} \\ $$$$=\begin{vmatrix}{{a}^{\mathrm{2}} −{b}^{\mathrm{2}} }&{{b}^{\mathrm{2}} −{c}^{\mathrm{2}} }&{{c}^{\mathrm{2}} }\\{{a}^{\mathrm{3}} −{b}^{\mathrm{3}} }&{{b}^{\mathrm{3}} −{c}^{\mathrm{3}} }&{{c}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{vmatrix}{by}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} −{C}_{\mathrm{2}} ,\:{C}_{\mathrm{2}} \rightarrow{C}_{\mathrm{2}} −{C}_{\mathrm{3}} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{{a}+{b}}&{{b}+{c}}&{{c}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} }&{{b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} }&{{c}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{vmatrix} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\begin{vmatrix}{{a}−{c}}&{{b}+{c}}&{{c}^{\mathrm{2}} }\\{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} +{ab}−{bc}}&{{b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} }&{{c}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{vmatrix}{by}\:{C}_{\mathrm{1}} \rightarrow{C}_{\mathrm{1}} −{C}_{\mathrm{2}} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({a}−{c}\right)\begin{vmatrix}{\mathrm{1}}&{{b}+{c}}&{{c}^{\mathrm{2}} }\\{{a}+{b}+{c}}&{{b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} }&{{c}^{\mathrm{3}} }\\{\mathrm{0}}&{\mathrm{0}}&{\mathrm{1}}\end{vmatrix} \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({a}−{c}\right)\left({b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} −{ab}−{ac}−{b}^{\mathrm{2}} −\mathrm{2}{bc}−{c}^{\mathrm{2}} \right) \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({a}−{c}\right)\left(−{bc}−{ab}−{ac}\right) \\ $$$$=\left({a}−{b}\right)\left({b}−{c}\right)\left({c}−{a}\right)\left({bc}+{ab}+{ac}\right) \\ $$
Commented by peter frank last updated on 06/Dec/18
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$