Question Number 11753 by Joel576 last updated on 31/Mar/17
$${a},\:{b},\:{c}\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{from}\:\mathrm{equation}\:{x}^{\mathrm{3}} \:−\:\mathrm{5}{x}^{\mathrm{2}} \:−\:\mathrm{9}{x}\:+\:\mathrm{10}\:=\:\mathrm{0} \\ $$$$\mathrm{If}\:{P}\left({x}\right)\:=\:{Ax}^{\mathrm{3}} \:+\:{Bx}^{\mathrm{2}} \:+\:{Cx}\:−\:\mathrm{2015}\:\mathrm{and} \\ $$$${P}\left({a}\right)\:=\:{b}\:+\:{c}\: \\ $$$${P}\left({b}\right)\:=\:{a}\:+\:{c} \\ $$$${P}\left({c}\right)\:=\:{a}\:+\:{b} \\ $$$$\mathrm{What}\:\mathrm{is}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{A}\:+\:{B}\:+\:{C}\:? \\ $$
Answered by sma3l2996 last updated on 31/Mar/17
$$\left({x}−{a}\right)\left({x}−{b}\right)\left({x}−{c}\right)={x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{10}=\mathrm{0} \\ $$$$\left({x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab}\right)\left({x}−{c}\right)={x}^{\mathrm{3}} −\left({a}+{b}+{c}\right){x}^{\mathrm{2}} +\left({ab}+{ac}+{bc}\right){x}−{abc} \\ $$$${a}+{b}+{c}=\mathrm{5}\:,\:{ab}+{ac}+{bc}=−\mathrm{9}\:,\:{abc}=−\mathrm{10} \\ $$$${we}\:{have}\:{P}\left({a}\right)+{P}\left({b}\right)+{P}\left({c}\right)=\mathrm{2}\left({a}+{b}+{c}\right)=\mathrm{2}×\mathrm{5} \\ $$$${P}\left({a}\right)+{P}\left({b}\right)+{P}\left({c}\right)={A}\left({a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} \right)+{B}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)+{C}\left({a}+{b}+{c}\right)−\mathrm{3}×\mathrm{2015} \\ $$$${we}\:{have}: \\ $$$$={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({a}^{\mathrm{2}} {b}+{a}^{\mathrm{2}} {c}+{b}^{\mathrm{2}} {c}+{bc}^{\mathrm{2}} +{ab}^{\mathrm{2}} +{ac}^{\mathrm{2}} \right)+\mathrm{6}{abc} \\ $$$${so}:\:{a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\left({a}+{b}+{c}\right)^{\mathrm{3}} −\mathrm{3}\left({a}\left({ab}+{ac}\right)+{b}\left({bc}+{ab}\right)+{c}\left({bc}+{ac}\right)\right)−\mathrm{6}{abc} \\ $$$$=\mathrm{5}^{\mathrm{3}} −\mathrm{3}\left({a}\left(−\mathrm{9}−{bc}\right)+{b}\left(−\mathrm{9}−{ac}\right)+{c}\left(−\mathrm{9}−{ab}\right)\right)−\mathrm{6}×\left(−\mathrm{10}\right) \\ $$$$=\mathrm{125}+\mathrm{3}\left(\mathrm{9}{a}+{abc}+\mathrm{9}{b}+{abc}+\mathrm{9}{c}+{abc}\right)+\mathrm{60} \\ $$$$=\mathrm{3}\left(\mathrm{9}\left({a}+{b}+{c}\right)+\mathrm{3}{abc}\right)+\mathrm{185} \\ $$$${a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} =\mathrm{27}×\mathrm{5}+\mathrm{9}×\left(−\mathrm{10}\right)+\mathrm{185}=\mathrm{230} \\ $$$${we}\:{have}\:{also} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{ac}+{bc}\right) \\ $$$${so}:\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{ac}+{bc}\right)=\mathrm{5}^{\mathrm{2}} −\mathrm{2}\left(−\mathrm{9}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\mathrm{43} \\ $$$${so}:\:{P}\left({a}\right)+{P}\left({b}\right)+{P}\left({c}\right)=\mathrm{230}{A}+\mathrm{43}{B}+\mathrm{5}{C}−\mathrm{6045}=\mathrm{2}×\mathrm{5}=\mathrm{10} \\ $$$$\mathrm{230}{A}+\mathrm{43}{B}+\mathrm{5}{C}=\mathrm{6055} \\ $$
Commented by sma3l2996 last updated on 31/Mar/17
$${working}... \\ $$
Commented by Joel576 last updated on 01/Apr/17
$${thank}\:{you}\:{very}\:{much} \\ $$
Answered by ajfour last updated on 31/Mar/17
$${A}+{B}+{C}\:=\:\mathrm{2625} \\ $$
Commented by ajfour last updated on 31/Mar/17
![As a,b, c are roots of x^3 −5x^2 −9x+10=0 or (x+2)(x^2 −7x+5) =0 are x=−2 , ((7±(√(29)))/2) we take a=−2; b+c=7; bc=((49−29)/4)= 5 ; a+b+c=5 . P(a)=b+c=Aa^3 +Ba^2 +Ca−2015 7=−8A+4B−2C−2015 .....(i) P(b)=a+c=Ab^3 +Bb^2 +Cb−2015 P(c)=a+b=Ac^3 +Bc^2 +Cc−2015 P(b)−P(c) gives b−c= (b−c)[A(b^2 +bc+c^2 ) +B(b+c)+C ] so, A[(b+c)^2 −bc] +B(b+c)+C=−1 ......(1) 44A+7B+C=−1 .....(ii) similarly [P(a)−P(b)]/(a−b) yields A(a^2 +ab+b^2 )+B(a+b)+C=−1 ......(2) and [P(c)−P(a)]/(c−a) gives A(c^2 +ac+a^2 )+B(c+a)+C=−1 ......(3) Adding eqns (1), (2), and (3) A{ 2(a^2 +b^2 +c^2 )+(ab+bc+ca)} +2B(a+b+c)+3C=−3 A{2(a+b+c)^2 −3(ab+bc+ca)} +2B(a+b+c)+3C=−3 or A{50−3(−9)}+10B+3C=−3 or 77A+10B+3C=−3 ...(iii) solving (i), (ii), and (iii) I found A+B+C = 2625 .](Q11785.png)
$${As}\:{a},{b},\:{c}\:{are}\:{roots}\:{of}\: \\ $$$${x}^{\mathrm{3}} −\mathrm{5}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{10}=\mathrm{0}\:\: \\ $$$${or}\:\:\left({x}+\mathrm{2}\right)\left({x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{5}\right)\:=\mathrm{0}\:\:\:{are} \\ $$$${x}=−\mathrm{2}\:,\:\frac{\mathrm{7}\pm\sqrt{\mathrm{29}}}{\mathrm{2}}\:\:{we}\:{take}\:{a}=−\mathrm{2}; \\ $$$${b}+{c}=\mathrm{7};\:{bc}=\frac{\mathrm{49}−\mathrm{29}}{\mathrm{4}}=\:\mathrm{5}\:;\: \\ $$$${a}+{b}+{c}=\mathrm{5}\:. \\ $$$${P}\left({a}\right)={b}+{c}={Aa}^{\mathrm{3}} +{Ba}^{\mathrm{2}} +{Ca}−\mathrm{2015} \\ $$$$\mathrm{7}=−\mathrm{8}{A}+\mathrm{4}{B}−\mathrm{2}{C}−\mathrm{2015}\:\:\:.....\left({i}\right) \\ $$$${P}\left({b}\right)={a}+{c}={Ab}^{\mathrm{3}} +{Bb}^{\mathrm{2}} +{Cb}−\mathrm{2015} \\ $$$${P}\left({c}\right)={a}+{b}={Ac}^{\mathrm{3}} +{Bc}^{\mathrm{2}} +{Cc}−\mathrm{2015} \\ $$$${P}\left({b}\right)−{P}\left({c}\right)\:{gives} \\ $$$${b}−{c}=\:\left({b}−{c}\right)\left[{A}\left({b}^{\mathrm{2}} +{bc}+{c}^{\mathrm{2}} \right)\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+{B}\left({b}+{c}\right)+{C}\:\right] \\ $$$${so},\:{A}\left[\left({b}+{c}\right)^{\mathrm{2}} −{bc}\right]\:+{B}\left({b}+{c}\right)+{C}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left(\mathrm{1}\right) \\ $$$$\mathrm{44}{A}+\mathrm{7}{B}+{C}=−\mathrm{1}\:\:.....\left({ii}\right) \\ $$$${similarly} \\ $$$$\left[{P}\left({a}\right)−{P}\left({b}\right)\right]/\left({a}−{b}\right)\:{yields} \\ $$$${A}\left({a}^{\mathrm{2}} +{ab}+{b}^{\mathrm{2}} \right)+{B}\left({a}+{b}\right)+{C}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left(\mathrm{2}\right) \\ $$$${and}\:\left[{P}\left({c}\right)−{P}\left({a}\right)\right]/\left({c}−{a}\right)\:{gives} \\ $$$${A}\left({c}^{\mathrm{2}} +{ac}+{a}^{\mathrm{2}} \right)+{B}\left({c}+{a}\right)+{C}=−\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:......\left(\mathrm{3}\right) \\ $$$${Adding}\:{eqns}\:\:\left(\mathrm{1}\right),\:\left(\mathrm{2}\right),\:{and}\:\left(\mathrm{3}\right) \\ $$$${A}\left\{\:\mathrm{2}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \:\right)+\left({ab}+{bc}+{ca}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{B}\left({a}+{b}+{c}\right)+\mathrm{3}{C}=−\mathrm{3} \\ $$$${A}\left\{\mathrm{2}\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{3}\left({ab}+{bc}+{ca}\right)\right\} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{B}\left({a}+{b}+{c}\right)+\mathrm{3}{C}=−\mathrm{3} \\ $$$${or}\: \\ $$$${A}\left\{\mathrm{50}−\mathrm{3}\left(−\mathrm{9}\right)\right\}+\mathrm{10}{B}+\mathrm{3}{C}=−\mathrm{3} \\ $$$${or}\:\:\:\:\:\mathrm{77}{A}+\mathrm{10}{B}+\mathrm{3}{C}=−\mathrm{3}\:\:\:...\left({iii}\right) \\ $$$${solving}\:\left({i}\right),\:\left({ii}\right),\:{and}\:\left({iii}\right) \\ $$$${I}\:{found}\:\:{A}+{B}+{C}\:=\:\mathrm{2625}\:. \\ $$
Commented by Joel576 last updated on 01/Apr/17
$${thank}\:{you}\:{very}\:{much} \\ $$