Question-38679

Question Number 38679 by Tinkutara last updated on 28/Jun/18

Answered by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

(a−x)(bc−bx−xc+x^2 −a^2 )−c(c^2 −cx−ab) +b(ac−b^2 +bx)=0 abc−abx−acx+ax^2 −a^3 −xbc+bx^2 +cx^2 −x^3 + a^2 x−c^3 +c^2 x+abc+abc−b^3 +b^2 x=0 3abc−a^3 −b^3 −c^3 +x(a^2 +b^2 +c^2 −ab−ac−bc)+ x^2 (a+b+c)−x^3 =0 since a+b+c=0 so 3abc−a^3 −b^3 −c^3 =0 x(a^2 +b^2 +c^2 −ab−bc−ac)−x^3 =0 now (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ac) 0=a^2 +b^2 +c^2 +2(ab+bc+ac) −ab−bc−ac=((a^2 +b^2 +c^2 )/2) so the eqn is x(a^2 +b^2 +c^2 +((a^2 +b^2 +c^2 )/2))−x^3 =0 hence either x=0 or (3/2)(a^2 +b^2 +c^2 )−x^2 =0 x=±(√((3/2)(a^2 +b^2 +c^2 ))) pleaze note element first row second collumn ( )_(1×2) is printed a but it shluld be c

$$\left({a}−{x}\right)\left({bc}−{bx}−{xc}+{x}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)−{c}\left({c}^{\mathrm{2}} −{cx}−{ab}\right) \\ $$$$+{b}\left({ac}−{b}^{\mathrm{2}} +{bx}\right)=\mathrm{0} \\ $$$${abc}−{abx}−{acx}+{ax}^{\mathrm{2}} −{a}^{\mathrm{3}} −{xbc}+{bx}^{\mathrm{2}} +{cx}^{\mathrm{2}} −{x}^{\mathrm{3}} + \\ $$$${a}^{\mathrm{2}} {x}−{c}^{\mathrm{3}} +{c}^{\mathrm{2}} {x}+{abc}+{abc}−{b}^{\mathrm{3}} +{b}^{\mathrm{2}} {x}=\mathrm{0} \\ $$$$\mathrm{3}{abc}−{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} +{x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{ac}−{bc}\right)+ \\ $$$${x}^{\mathrm{2}} \left({a}+{b}+{c}\right)−{x}^{\mathrm{3}} =\mathrm{0} \\ $$$${since}\:{a}+{b}+{c}=\mathrm{0}\:{so}\:\mathrm{3}{abc}−{a}^{\mathrm{3}} −{b}^{\mathrm{3}} −{c}^{\mathrm{3}} =\mathrm{0} \\ $$$${x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} −{ab}−{bc}−{ac}\right)−{x}^{\mathrm{3}} =\mathrm{0} \\ $$$${now}\:\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ac}\right) \\ $$$$\mathrm{0}={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ac}\right) \\ $$$$−{ab}−{bc}−{ac}=\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}} \\ $$$${so}\:{the}\:{eqn}\:{is} \\ $$$${x}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\frac{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} }{\mathrm{2}}\right)−{x}^{\mathrm{3}} =\mathrm{0} \\ $$$${hence}\:{either}\:{x}=\mathrm{0}\:{or} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)−{x}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\pm\sqrt{\frac{\mathrm{3}}{\mathrm{2}}\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} \right)}\: \\ $$$${pleaze}\:{note}\:{element}\:{first}\:{row}\:{second}\:{collumn} \\ $$$$\left(\:\right)_{\mathrm{1}×\mathrm{2}} \:{is}\:{printed}\:{a}\:{but}\:{it}\:{shluld}\:{be}\:{c} \\ $$$$ \\ $$

Commented by Tinkutara last updated on 28/Jun/18

Shouldn't we use properties of determinants to get a factorized expression?

Commented by tanmay.chaudhury50@gmail.com last updated on 28/Jun/18

$${yes}..{let}\:{me}\:{solve}\:{by}\:{another}\:{method}…{using}\:{properties} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18

$${pls}\:{visit}\:{archive}.{org}\:{and}\:{download}\: \\ $$$${mathematical}\:{formula}\:{and}\:{tables}…{others} \\ $$$${books}…{murray}\:{spiegel}…. \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 04/Jul/18