Question Number 222225 by mathBagja last updated on 21/Jun/25
$${a}+{b}+{c}=−{D} \\ $$$${ab}+{bc}+{ac}={E} \\ $$$${abc}=−\mathrm{1} \\ $$$$\left({a}^{\mathrm{2}} −{bc}\right)\left({b}^{\mathrm{2}} −{ac}\right)\left({c}^{\mathrm{2}} −{ab}\right)=\mathrm{40} \\ $$$${D}−{E}=\mathrm{4} \\ $$$$\left({D}+{E}\right)^{\mathrm{2}} =… \\ $$
Commented by mr W last updated on 20/Jun/25
$${there}\:{is}\:{no}\:{unique}\:{value}\:{for}\:\left({D}+{E}\right)^{\mathrm{2}} . \\ $$$${please}\:{recheck}\:{the}\:{question}! \\ $$
Commented by fantastic last updated on 20/Jun/25
$${is}\:{the}\:{answer}\:{is}\:{a}\:{number}?? \\ $$
Commented by fantastic last updated on 20/Jun/25
$${me}\:{too}\:{got}\:{this} \\ $$
Commented by mr W last updated on 20/Jun/25
$${i}\:{only}\:{got}\:{D}^{\mathrm{3}} −{E}^{\mathrm{3}} =\mathrm{40}. \\ $$
Commented by A5T last updated on 20/Jun/25
![I also got D^3 −E^3 =40 (a^2 −((abc)/a))(b^2 −((abc)/b))(c^2 −((abc)/c))=40 ⇒(a^2 +(1/a))(b^2 +(1/b))(c^2 +(1/c))=40 ⇒(a^3 +1)(b^3 +1)(c^3 +1)=40abc=−40 ⇒(a^3 b^3 c^3 +a^3 b^3 +a^3 c^3 +a^3 +b^3 c^3 +b^3 +c^3 +1)=−40 ⇒a^3 +b^3 +c^3 +a^3 b^3 +a^3 c^3 +b^3 c^3 =−40 a^3 +b^3 +c^3 −3abc=(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca) =−D[D^2 −3E] Similarly, a^3 b^3 +b^3 c^3 +c^3 a^3 −3(abc)^2 =E^3 −3ED ⇒a^3 +b^3 +c^3 +a^3 b^3 +a^3 c^3 +b^3 c^3 =E^3 −3ED−D^3 +3ED+3−3 ⇒E^3 −D^3 =−40](https://www.tinkutara.com/question/Q222247.png)
$$\mathrm{I}\:\mathrm{also}\:\mathrm{got}\:\mathrm{D}^{\mathrm{3}} −\mathrm{E}^{\mathrm{3}} =\mathrm{40} \\ $$$$ \\ $$$$\left(\mathrm{a}^{\mathrm{2}} −\frac{\mathrm{abc}}{\mathrm{a}}\right)\left(\mathrm{b}^{\mathrm{2}} −\frac{\mathrm{abc}}{\mathrm{b}}\right)\left(\mathrm{c}^{\mathrm{2}} −\frac{\mathrm{abc}}{\mathrm{c}}\right)=\mathrm{40} \\ $$$$\Rightarrow\left(\mathrm{a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{a}}\right)\left(\mathrm{b}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{b}}\right)\left(\mathrm{c}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{c}}\right)=\mathrm{40} \\ $$$$\Rightarrow\left(\mathrm{a}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{b}^{\mathrm{3}} +\mathrm{1}\right)\left(\mathrm{c}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{40abc}=−\mathrm{40} \\ $$$$\Rightarrow\left(\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{1}\right)=−\mathrm{40} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} =−\mathrm{40} \\ $$$$\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} −\mathrm{3abc}=\left(\mathrm{a}+\mathrm{b}+\mathrm{c}\right)\left(\mathrm{a}^{\mathrm{2}} +\mathrm{b}^{\mathrm{2}} +\mathrm{c}^{\mathrm{2}} −\mathrm{ab}−\mathrm{bc}−\mathrm{ca}\right) \\ $$$$=−\mathrm{D}\left[\mathrm{D}^{\mathrm{2}} −\mathrm{3E}\right] \\ $$$$\mathrm{Similarly},\:\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} \mathrm{a}^{\mathrm{3}} −\mathrm{3}\left(\mathrm{abc}\right)^{\mathrm{2}} =\mathrm{E}^{\mathrm{3}} −\mathrm{3ED} \\ $$$$\Rightarrow\mathrm{a}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} +\mathrm{c}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{b}^{\mathrm{3}} +\mathrm{a}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} +\mathrm{b}^{\mathrm{3}} \mathrm{c}^{\mathrm{3}} =\mathrm{E}^{\mathrm{3}} −\mathrm{3ED}−\mathrm{D}^{\mathrm{3}} +\mathrm{3ED}+\mathrm{3}−\mathrm{3} \\ $$$$\Rightarrow\mathrm{E}^{\mathrm{3}} −\mathrm{D}^{\mathrm{3}} =−\mathrm{40} \\ $$
Commented by mathBagja last updated on 21/Jun/25
$${I}'{m}\:{sorry}\:{Sir},\:{I}\:{Missed}\:{Something} \\ $$
Commented by mathBagja last updated on 21/Jun/25
$${yes}\:{Sir} \\ $$
Commented by fantastic last updated on 21/Jun/25
$${I}\:{also}\:{got}\:{that} \\ $$
Commented by fantastic last updated on 21/Jun/25
![(a^2 −bc)(b^2 −ac)(c^2 −ab) will be =−[(ab+bc+ca)((ab)^2 +(bc)^2 +(ca)^2 −(ab^2 c+abc^2 +a^2 bc))+3]− [(a+b+c)(a^2 +b^2 +c^2 −ab−bc−ca)−3] =−[E(E^2 −3D)+3]+D(D^2 −3E)+3 So −E^3 +3ED−3+D^3 −3ED+3=40 ∴D^3 −E^3 =40 But we can not derive (D+E)^2 from this](https://www.tinkutara.com/question/Q222255.png)
$$\left({a}^{\mathrm{2}} −{bc}\right)\left({b}^{\mathrm{2}} −{ac}\right)\left({c}^{\mathrm{2}} −{ab}\right)\:{will}\:{be} \\ $$$$=−\left[\left({ab}+{bc}+{ca}\right)\left(\left({ab}\right)^{\mathrm{2}} +\left({bc}\right)^{\mathrm{2}} +\left({ca}\right)^{\mathrm{2}} −\left({ab}^{\mathrm{2}} {c}+{abc}^{\mathrm{2}} +{a}^{\mathrm{2}} {bc}\right)\right)+\mathrm{3}\right]− \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[\left({a}+{b}+{c}\right)\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ca}\right)−\mathrm{3}\right] \\ $$$$=−\left[{E}\left({E}^{\mathrm{2}} −\mathrm{3}{D}\right)+\mathrm{3}\right]+{D}\left({D}^{\mathrm{2}} −\mathrm{3}{E}\right)+\mathrm{3} \\ $$$${So}\:−{E}^{\mathrm{3}} \cancel{+\mathrm{3}{ED}−\mathrm{3}}+{D}^{\mathrm{3}} \cancel{−\mathrm{3}{ED}+\mathrm{3}}=\mathrm{40} \\ $$$$\therefore{D}^{\mathrm{3}} −{E}^{\mathrm{3}} =\mathrm{40} \\ $$$${But}\:{we}\:{can}\:{not}\:{derive}\:\left({D}+{E}\right)^{\mathrm{2}} \:{from}\:{this} \\ $$