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Question Number 199092 by necx122 last updated on 27/Oct/23
Let the polynomial p(x)=5x^3 +3x^2 −10 have roots a,b and c. What is the value of (a/(b+c))+(b/(c+a))+(c/(a+b))?
$${Let}\:{the}\:{polynomial}\:{p}\left({x}\right)=\mathrm{5}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10} \\ $$$${have}\:{roots}\:{a},{b}\:{and}\:{c}.\:{What}\:{is}\:{the}\:{value} \\ $$$${of}\:\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}? \\ $$
Answered by Rasheed.Sindhi last updated on 28/Oct/23
p(x)=5x^3 +3x^2 −10 a+b+c=−(3/5) , ab+bc+ca=0 , abc=2 a^2 +b^2 +c^2 =(a+b+c)^2 −2(ab+bc+ca) a^2 +b^2 +c^2 =(−(3/5))^2 −2(0)=(9/(25)) (a/(b+c))+1+(b/(c+a))+1+(c/(a+b))+1−3 =((a+b+c)/(b+c))+((b+c+a)/(c+a))+((c+a+b)/(a+b))−3 =(a+b+c)((1/(b+c))+(1/(c+a))+(1/(a+b)))−3 =−(3/5)((1/(−(3/5)−a))+(1/(−(3/5)−b))+(1/(−(3/5)−c)))−3 =−(3/5)((5/(−3−5a))+(5/(−3−5b))+(5/(−3−5c)))−3 =3((1/(3+5a))+(1/(3+5b))+(1/(3+5c)))−3 =3((((3+5a)(3+5b)+(3+5b)(3+5c)+(3+5a)(3+5c))/((3+5a)(3+5b)(3+5c))))−3 =3(((9+15a+15b+25ab+9+15b+15c+25bc+9+15c+15a+25ca)/(27+45(a+b+c)+75(ab+bc+ca)+125abc)))−3 =3(((27+30(a+b+c)+25(ab+bc+ca))/(27+45(a+b+c)+75(ab+bc+ca)+125abc)))−3 =3(((27+30(−(3/5))+25(0))/(27+45(−(3/5))+75(0)+125(2))))−3 =3(((27−18)/(27−27+250)))−3=((27−750)/(250))=−((723)/(250))
$${p}\left({x}\right)=\mathrm{5}{x}^{\mathrm{3}} +\mathrm{3}{x}^{\mathrm{2}} −\mathrm{10} \\ $$$${a}+{b}+{c}=−\frac{\mathrm{3}}{\mathrm{5}}\:,\:{ab}+{bc}+{ca}=\mathrm{0}\:,\:{abc}=\mathrm{2} \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left({a}+{b}+{c}\right)^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$${a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} =\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{0}\right)=\frac{\mathrm{9}}{\mathrm{25}} \\ $$$$\:\frac{{a}}{{b}+{c}}+\mathrm{1}+\frac{{b}}{{c}+{a}}+\mathrm{1}+\frac{{c}}{{a}+{b}}+\mathrm{1}−\mathrm{3} \\ $$$$=\frac{{a}+{b}+{c}}{{b}+{c}}+\frac{{b}+{c}+{a}}{{c}+{a}}+\frac{{c}+{a}+{b}}{{a}+{b}}−\mathrm{3} \\ $$$$=\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{a}+{b}}\right)−\mathrm{3} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{5}}\left(\frac{\mathrm{1}}{−\frac{\mathrm{3}}{\mathrm{5}}−{a}}+\frac{\mathrm{1}}{−\frac{\mathrm{3}}{\mathrm{5}}−{b}}+\frac{\mathrm{1}}{−\frac{\mathrm{3}}{\mathrm{5}}−{c}}\right)−\mathrm{3} \\ $$$$=−\frac{\mathrm{3}}{\mathrm{5}}\left(\frac{\mathrm{5}}{−\mathrm{3}−\mathrm{5}{a}}+\frac{\mathrm{5}}{−\mathrm{3}−\mathrm{5}{b}}+\frac{\mathrm{5}}{−\mathrm{3}−\mathrm{5}{c}}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{1}}{\mathrm{3}+\mathrm{5}{a}}+\frac{\mathrm{1}}{\mathrm{3}+\mathrm{5}{b}}+\frac{\mathrm{1}}{\mathrm{3}+\mathrm{5}{c}}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\left(\mathrm{3}+\mathrm{5}{a}\right)\left(\mathrm{3}+\mathrm{5}{b}\right)+\left(\mathrm{3}+\mathrm{5}{b}\right)\left(\mathrm{3}+\mathrm{5}{c}\right)+\left(\mathrm{3}+\mathrm{5}{a}\right)\left(\mathrm{3}+\mathrm{5}{c}\right)}{\left(\mathrm{3}+\mathrm{5}{a}\right)\left(\mathrm{3}+\mathrm{5}{b}\right)\left(\mathrm{3}+\mathrm{5}{c}\right)}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{9}+\mathrm{15}{a}+\mathrm{15}{b}+\mathrm{25}{ab}+\mathrm{9}+\mathrm{15}{b}+\mathrm{15}{c}+\mathrm{25}{bc}+\mathrm{9}+\mathrm{15}{c}+\mathrm{15}{a}+\mathrm{25}{ca}}{\mathrm{27}+\mathrm{45}\left({a}+{b}+{c}\right)+\mathrm{75}\left({ab}+{bc}+{ca}\right)+\mathrm{125}{abc}}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{27}+\mathrm{30}\left({a}+{b}+{c}\right)+\mathrm{25}\left({ab}+{bc}+{ca}\right)}{\mathrm{27}+\mathrm{45}\left({a}+{b}+{c}\right)+\mathrm{75}\left({ab}+{bc}+{ca}\right)+\mathrm{125}{abc}}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{27}+\mathrm{30}\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{25}\left(\mathrm{0}\right)}{\mathrm{27}+\mathrm{45}\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{75}\left(\mathrm{0}\right)+\mathrm{125}\left(\mathrm{2}\right)}\right)−\mathrm{3} \\ $$$$=\mathrm{3}\left(\frac{\mathrm{27}−\mathrm{18}}{\mathrm{27}−\mathrm{27}+\mathrm{250}}\right)−\mathrm{3}=\frac{\mathrm{27}−\mathrm{750}}{\mathrm{250}}=−\frac{\mathrm{723}}{\mathrm{250}} \\ $$
Commented by AST last updated on 28/Oct/23
This method may not always work when b,c ∈C\R
$${This}\:{method}\:{may}\:{not}\:{always}\:{work}\:{when}\:{b},{c}\:\in\mathbb{C}\backslash\mathbb{R} \\ $$
Commented by Rasheed.Sindhi last updated on 28/Oct/23
You′re very right AST sir! ThanX!
$${You}'{re}\:{very}\:{right}\:{AST}\:{sir}! \\ $$$$\boldsymbol{\mathcal{T}{han}\mathcal{X}}! \\ $$
Commented by mr W last updated on 28/Oct/23
Rasheed sir: i think your path is correct. but take care in last step: abc=2 ≠10, then you also get −((723)/(250)).
$${Rasheed}\:{sir}: \\ $$$${i}\:{think}\:{your}\:{path}\:{is}\:{correct}.\:{but}\:{take} \\ $$$${care}\:{in}\:{last}\:{step}:\:{abc}=\mathrm{2}\:\neq\mathrm{10},\:{then} \\ $$$${you}\:{also}\:{get}\:−\frac{\mathrm{723}}{\mathrm{250}}. \\ $$
Commented by Rasheed.Sindhi last updated on 28/Oct/23
A lot of Thαnks sir!
$${A}\:{lot}\:{of}\:\mathcal{T}{h}\alpha{nks}\:\boldsymbol{{sir}}! \\ $$
Answered by AST last updated on 28/Oct/23
Let a=a;b=x+yi,c=e+fi a+b+c=((−3)/5)⇒f=−y(otherwise,sum won′t ∈R) abc=2⇒a(x+yi)(e−yi)=2⇒e=x ⇒c=x−yi⇒abc=a(x^2 +y^2 )=2;a+2x=((−3)/5); ab+bc+ca=0⇒a(x+yi)+x^2 +y^2 +a(x−yi)=0 ⇒2ax+x^2 +y^2 =0⇒a(−2ax)=2⇒a^2 x=−1 (a/(b+c))+(b/(c+a))+(c/(a+b))=(a+b+c)((1/(b+c))+(1/(c+a))+(1/(a+b)))−3 =(a+2x)((1/(2x))+((2(x+a))/(x^2 +a^2 +2ax+y^2 )))−3=(((−3)/5))((1/(2x))+((2(x+a))/a^2 ))−3 =(((−3)/5))(((a^2 +4x^2 +4ax)/(2a^2 x)))−3=(((−3)/5))(((a+2x)^2 )/(2a^2 x))−3 =(((−3)/5))((((9/(25))))/((−2)/1))−3=((9/(−50))×((−3)/5))−3=((27)/(250))−3=−((723)/(250))
$${Let}\:{a}={a};{b}={x}+{yi},{c}={e}+{fi} \\ $$$${a}+{b}+{c}=\frac{−\mathrm{3}}{\mathrm{5}}\Rightarrow{f}=−{y}\left({otherwise},{sum}\:{won}'{t}\:\in\mathbb{R}\right) \\ $$$${abc}=\mathrm{2}\Rightarrow{a}\left({x}+{yi}\right)\left({e}−{yi}\right)=\mathrm{2}\Rightarrow{e}={x} \\ $$$$\Rightarrow{c}={x}−{yi}\Rightarrow{abc}={a}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)=\mathrm{2};{a}+\mathrm{2}{x}=\frac{−\mathrm{3}}{\mathrm{5}}; \\ $$$${ab}+{bc}+{ca}=\mathrm{0}\Rightarrow{a}\left({x}+{yi}\right)+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{a}\left({x}−{yi}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{ax}+{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{0}\Rightarrow{a}\left(−\mathrm{2}{ax}\right)=\mathrm{2}\Rightarrow{a}^{\mathrm{2}} {x}=−\mathrm{1} \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}=\left({a}+{b}+{c}\right)\left(\frac{\mathrm{1}}{{b}+{c}}+\frac{\mathrm{1}}{{c}+{a}}+\frac{\mathrm{1}}{{a}+{b}}\right)−\mathrm{3} \\ $$$$=\left({a}+\mathrm{2}{x}\right)\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{2}\left({x}+{a}\right)}{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} +\mathrm{2}{ax}+{y}^{\mathrm{2}} }\right)−\mathrm{3}=\left(\frac{−\mathrm{3}}{\mathrm{5}}\right)\left(\frac{\mathrm{1}}{\mathrm{2}{x}}+\frac{\mathrm{2}\left({x}+{a}\right)}{{a}^{\mathrm{2}} }\right)−\mathrm{3} \\ $$$$=\left(\frac{−\mathrm{3}}{\mathrm{5}}\right)\left(\frac{{a}^{\mathrm{2}} +\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{ax}}{\mathrm{2}{a}^{\mathrm{2}} {x}}\right)−\mathrm{3}=\left(\frac{−\mathrm{3}}{\mathrm{5}}\right)\frac{\left({a}+\mathrm{2}{x}\right)^{\mathrm{2}} }{\mathrm{2}{a}^{\mathrm{2}} {x}}−\mathrm{3} \\ $$$$=\left(\frac{−\mathrm{3}}{\mathrm{5}}\right)\frac{\left(\frac{\mathrm{9}}{\mathrm{25}}\right)}{\frac{−\mathrm{2}}{\mathrm{1}}}−\mathrm{3}=\left(\frac{\mathrm{9}}{−\mathrm{50}}×\frac{−\mathrm{3}}{\mathrm{5}}\right)−\mathrm{3}=\frac{\mathrm{27}}{\mathrm{250}}−\mathrm{3}=−\frac{\mathrm{723}}{\mathrm{250}} \\ $$
Answered by mr W last updated on 28/Oct/23
Method I let s=a+b+c=−(3/5) ab+bc+ca=0 abc=((10)/5)=2 (a/(b+c))+(b/(c+a))+(c/(a+b)) =(a/(s−a))+(b/(s−b))+(c/(s−c)) =(((a+b+c)s^2 −2(ab+bc+ca)s+3abc)/((s−a)(s−b)(s−c))) =(((a+b+c)s^2 −2(ab+bc+ca)s+3abc)/(s^3 −(a+b+c)s^2 +(ab+bc+ca)s−abc)) =((s^3 −2(ab+bc+ca)s+3abc)/((ab+bc+ca)s−abc)) =((s^3 +3abc)/(−abc)) =−(s^3 /(abc))−3 =−(−(3/5))^3 ×(1/2)−3=−((723)/(250)) ✓
$$\underline{\boldsymbol{{Method}}\:\boldsymbol{{I}}} \\ $$$${let}\:{s}={a}+{b}+{c}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${ab}+{bc}+{ca}=\mathrm{0} \\ $$$${abc}=\frac{\mathrm{10}}{\mathrm{5}}=\mathrm{2} \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}} \\ $$$$=\frac{{a}}{{s}−{a}}+\frac{{b}}{{s}−{b}}+\frac{{c}}{{s}−{c}} \\ $$$$=\frac{\left({a}+{b}+{c}\right){s}^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right){s}+\mathrm{3}{abc}}{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)} \\ $$$$=\frac{\left({a}+{b}+{c}\right){s}^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right){s}+\mathrm{3}{abc}}{{s}^{\mathrm{3}} −\left({a}+{b}+{c}\right){s}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){s}−{abc}} \\ $$$$=\frac{{s}^{\mathrm{3}} −\mathrm{2}\left({ab}+{bc}+{ca}\right){s}+\mathrm{3}{abc}}{\left({ab}+{bc}+{ca}\right){s}−{abc}} \\ $$$$=\frac{{s}^{\mathrm{3}} +\mathrm{3}{abc}}{−{abc}} \\ $$$$=−\frac{{s}^{\mathrm{3}} }{{abc}}−\mathrm{3} \\ $$$$=−\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{3}} ×\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{3}=−\frac{\mathrm{723}}{\mathrm{250}}\:\checkmark \\ $$
Commented by mr W last updated on 28/Oct/23
Method II let s=a+b+c=−(3/5) let u=(x/(s−x)), i.e. x=((su)/(u+1)) 5(((su)/(u+1)))^3 +3(((su)/(u+1)))^2 −10=0 5(su)^3 +3(su)^2 (u+1)−10(u+1)^3 =0 (5s^3 +3s^2 −10)u^3 +3(s^2 −10)u^2 −30u−10=0 (a/(b+c))+(b/(c+a))+(c/(a+b))=(a/(s−a))+(b/(s−b))+(c/(s−c)) =(x_1 /(s−x_1 ))+(x_2 /(s−x_2 ))+(x_3 /(s−x_3 ))=u_1 +u_2 +u_3 =−((3(s^2 −10))/(5s^3 +3s^2 −10))=−((723)/(250)) ✓
$$\underline{\boldsymbol{{Method}}\:\boldsymbol{{II}}} \\ $$$${let}\:{s}={a}+{b}+{c}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${let}\:{u}=\frac{{x}}{{s}−{x}},\:{i}.{e}.\:{x}=\frac{{su}}{{u}+\mathrm{1}} \\ $$$$\mathrm{5}\left(\frac{{su}}{{u}+\mathrm{1}}\right)^{\mathrm{3}} +\mathrm{3}\left(\frac{{su}}{{u}+\mathrm{1}}\right)^{\mathrm{2}} −\mathrm{10}=\mathrm{0} \\ $$$$\mathrm{5}\left({su}\right)^{\mathrm{3}} +\mathrm{3}\left({su}\right)^{\mathrm{2}} \left({u}+\mathrm{1}\right)−\mathrm{10}\left({u}+\mathrm{1}\right)^{\mathrm{3}} =\mathrm{0} \\ $$$$\left(\mathrm{5}{s}^{\mathrm{3}} +\mathrm{3}{s}^{\mathrm{2}} −\mathrm{10}\right){u}^{\mathrm{3}} +\mathrm{3}\left({s}^{\mathrm{2}} −\mathrm{10}\right){u}^{\mathrm{2}} −\mathrm{30}{u}−\mathrm{10}=\mathrm{0} \\ $$$$ \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}}=\frac{{a}}{{s}−{a}}+\frac{{b}}{{s}−{b}}+\frac{{c}}{{s}−{c}} \\ $$$$=\frac{{x}_{\mathrm{1}} }{{s}−{x}_{\mathrm{1}} }+\frac{{x}_{\mathrm{2}} }{{s}−{x}_{\mathrm{2}} }+\frac{{x}_{\mathrm{3}} }{{s}−{x}_{\mathrm{3}} }={u}_{\mathrm{1}} +{u}_{\mathrm{2}} +{u}_{\mathrm{3}} \\ $$$$\:\:=−\frac{\mathrm{3}\left({s}^{\mathrm{2}} −\mathrm{10}\right)}{\mathrm{5}{s}^{\mathrm{3}} +\mathrm{3}{s}^{\mathrm{2}} −\mathrm{10}}=−\frac{\mathrm{723}}{\mathrm{250}}\:\checkmark \\ $$
Commented by mr W last updated on 28/Oct/23
Method III s=a+b+c=−(3/5) ab+bc+ca=0 abc=((10)/5)=2 (a/(b+c))+(b/(c+a))+(c/(a+b)) =((a+b+c)/(b+c))+((a+b+c)/(c+a))+((a+b+c)/(a+b))−3 =s((1/(s−a))+(1/(s−b))+(1/(s−c)))−3 =s×((3s^2 −2(a+b+c)s+ab+bc+ca)/((s−a)(s−b)(s−c)))−3 =s×((3s^2 −2s^2 +ab+bc+ca)/(s^3 −(a+b+c)s^2 +(ab+bc+ca)s−abc))−3 =s×((s^2 +ab+bc+ca)/((ab+bc+ca)s−abc))−3 =(s^3 /(−abc))−3 =−(1/2)(−(3/5))^3 −3=−((723)/(250)) ✓
$$\boldsymbol{{Method}}\:\boldsymbol{{III}} \\ $$$${s}={a}+{b}+{c}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${ab}+{bc}+{ca}=\mathrm{0} \\ $$$${abc}=\frac{\mathrm{10}}{\mathrm{5}}=\mathrm{2} \\ $$$$\frac{{a}}{{b}+{c}}+\frac{{b}}{{c}+{a}}+\frac{{c}}{{a}+{b}} \\ $$$$=\frac{{a}+{b}+{c}}{{b}+{c}}+\frac{{a}+{b}+{c}}{{c}+{a}}+\frac{{a}+{b}+{c}}{{a}+{b}}−\mathrm{3} \\ $$$$={s}\left(\frac{\mathrm{1}}{{s}−{a}}+\frac{\mathrm{1}}{{s}−{b}}+\frac{\mathrm{1}}{{s}−{c}}\right)−\mathrm{3} \\ $$$$={s}×\frac{\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}\left({a}+{b}+{c}\right){s}+{ab}+{bc}+{ca}}{\left({s}−{a}\right)\left({s}−{b}\right)\left({s}−{c}\right)}−\mathrm{3} \\ $$$$={s}×\frac{\mathrm{3}{s}^{\mathrm{2}} −\mathrm{2}{s}^{\mathrm{2}} +{ab}+{bc}+{ca}}{{s}^{\mathrm{3}} −\left({a}+{b}+{c}\right){s}^{\mathrm{2}} +\left({ab}+{bc}+{ca}\right){s}−{abc}}−\mathrm{3} \\ $$$$={s}×\frac{{s}^{\mathrm{2}} +{ab}+{bc}+{ca}}{\left({ab}+{bc}+{ca}\right){s}−{abc}}−\mathrm{3} \\ $$$$=\frac{{s}^{\mathrm{3}} }{−{abc}}−\mathrm{3} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\left(−\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{3}} −\mathrm{3}=−\frac{\mathrm{723}}{\mathrm{250}}\:\checkmark \\ $$
Commented by Rasheed.Sindhi last updated on 28/Oct/23
Great sir!
$$\mathbb{G}\boldsymbol{\mathrm{reat}}\:\boldsymbol{\mathrm{sir}}! \\ $$
Commented by necx122 last updated on 28/Oct/23
This is one thing I love about this platform, the seriousness in giving us answers that are accurate and the time spent in offering this untamed service to us. Personally, I'm grateful sirs for this several methods; I understand them. @Rashes.Sindhi and @Mr. W
Answered by ajfour last updated on 28/Oct/23
a+b+c=−(3/5)=s ab+bc+ca=0 abc=2 b+c=−((bc)/a)=−(2/a^2 ) Q=Σ((s−(b+c))/(b+c)) (Q+3)(−(5/3))=Σ(1/(b+c)) ((5Q)/3)+5=(1/2)Σa^2 ((5Q)/3)+5=(1/2){s^2 −2(ab+bc+ca)} 5((Q/3)+1)=(1/2)(s^2 −0)=(9/(50)) Q=3((9/(250))−1)=−((723)/(250))
$${a}+{b}+{c}=−\frac{\mathrm{3}}{\mathrm{5}}={s} \\ $$$${ab}+{bc}+{ca}=\mathrm{0} \\ $$$${abc}=\mathrm{2} \\ $$$${b}+{c}=−\frac{{bc}}{{a}}=−\frac{\mathrm{2}}{{a}^{\mathrm{2}} } \\ $$$${Q}=\Sigma\frac{{s}−\left({b}+{c}\right)}{{b}+{c}} \\ $$$$\left({Q}+\mathrm{3}\right)\left(−\frac{\mathrm{5}}{\mathrm{3}}\right)=\Sigma\frac{\mathrm{1}}{{b}+{c}} \\ $$$$\frac{\mathrm{5}{Q}}{\mathrm{3}}+\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}\Sigma{a}^{\mathrm{2}} \\ $$$$\frac{\mathrm{5}{Q}}{\mathrm{3}}+\mathrm{5}=\frac{\mathrm{1}}{\mathrm{2}}\left\{{s}^{\mathrm{2}} −\mathrm{2}\left({ab}+{bc}+{ca}\right)\right\}\: \\ $$$$\:\mathrm{5}\left(\frac{{Q}}{\mathrm{3}}+\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left({s}^{\mathrm{2}} −\mathrm{0}\right)=\frac{\mathrm{9}}{\mathrm{50}} \\ $$$${Q}=\mathrm{3}\left(\frac{\mathrm{9}}{\mathrm{250}}−\mathrm{1}\right)=−\frac{\mathrm{723}}{\mathrm{250}} \\ $$
Commented by necx122 last updated on 28/Oct/23
just when I was in awe of the amount of steps used in solving the question here is Mr. Ajfour with another mind blowing approach. I'm grateful sir. Thank you.
Answered by Frix last updated on 28/Oct/23
x^3 +px^2 +qx+r=0 (x_1 /(x_2 +x_3 ))+(x_2 /(x_1 +x_3 ))+(x_3 /(x_1 +x_2 ))=−2+((p^3 +r)/(pq−r))
$${x}^{\mathrm{3}} +{px}^{\mathrm{2}} +{qx}+{r}=\mathrm{0} \\ $$$$\frac{{x}_{\mathrm{1}} }{{x}_{\mathrm{2}} +{x}_{\mathrm{3}} }+\frac{{x}_{\mathrm{2}} }{{x}_{\mathrm{1}} +{x}_{\mathrm{3}} }+\frac{{x}_{\mathrm{3}} }{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }=−\mathrm{2}+\frac{{p}^{\mathrm{3}} +{r}}{{pq}−{r}} \\ $$

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