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Question Number 25066 by Mr easy last updated on 02/Dec/17
let a,b,c,x,y and z be complex number such that a=((b+c)/(x−2)) ,b=((c+a)/(y−2)) c=((a+b)/(z−2)). xy +yz +zx=1000 and x+y+z=2016 find the value of xyz.
$${let}\:{a},{b},{c},{x},{y}\:{and}\:{z}\:{be}\:{complex}\:{number} \\ $$$${such}\:{that}\:{a}=\frac{{b}+{c}}{{x}−\mathrm{2}}\:,{b}=\frac{{c}+{a}}{{y}−\mathrm{2}}\:\:\:\:{c}=\frac{{a}+{b}}{{z}−\mathrm{2}}. \\ $$$${xy}\:+{yz}\:+{zx}=\mathrm{1000}\:{and}\:{x}+{y}+{z}=\mathrm{2016} \\ $$$${find}\:{the}\:{value}\:{of}\:{xyz}. \\ $$
Answered by ajfour last updated on 03/Dec/17
Π(x−2)=xyz−2Σxy+4Σx−8 (((a+b)(b+c)(c+a))/(abc))=xyz−2000 +8064−8 ⇒ xyz=(((a+b)(b+c)(c+a))/(abc))−6056 Further, x+y+z=8+Σ(((b+c)/a))=2016 ...(i) Σ(x−2)(y−2)=Σ(((b+c)/a))(((c+a)/b)) Σxy−2Σ(x+y)+12=Σ(((b+c)/a))(((c+a)/b)) 1000−4×2016+12=Σ(((b+c)/a))(((c+a)/b)) ⇒ Σ(((b+c)/a))(((c+a)/b))=7056 so ((b+c)/a), ((c+a)/b), ((a+b)/c) are roots of cubic equation: q^3 −2008q^2 +7056q−(xyz+6056)=0 ....may continue..
$$\Pi\left({x}−\mathrm{2}\right)={xyz}−\mathrm{2}\Sigma{xy}+\mathrm{4}\Sigma{x}−\mathrm{8} \\ $$$$\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{{abc}}={xyz}−\mathrm{2000} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{8064}−\mathrm{8} \\ $$$$\Rightarrow\:{xyz}=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{{abc}}−\mathrm{6056} \\ $$$${Further}, \\ $$$${x}+{y}+{z}=\mathrm{8}+\Sigma\left(\frac{{b}+{c}}{{a}}\right)=\mathrm{2016}\:\:…\left({i}\right) \\ $$$$\Sigma\left({x}−\mathrm{2}\right)\left({y}−\mathrm{2}\right)=\Sigma\left(\frac{{b}+{c}}{{a}}\right)\left(\frac{{c}+{a}}{{b}}\right) \\ $$$$\Sigma{xy}−\mathrm{2}\Sigma\left({x}+{y}\right)+\mathrm{12}=\Sigma\left(\frac{{b}+{c}}{{a}}\right)\left(\frac{{c}+{a}}{{b}}\right) \\ $$$$\mathrm{1000}−\mathrm{4}×\mathrm{2016}+\mathrm{12}=\Sigma\left(\frac{{b}+{c}}{{a}}\right)\left(\frac{{c}+{a}}{{b}}\right) \\ $$$$\Rightarrow\:\Sigma\left(\frac{{b}+{c}}{{a}}\right)\left(\frac{{c}+{a}}{{b}}\right)=\mathrm{7056} \\ $$$${so}\:\frac{{b}+{c}}{{a}},\:\frac{{c}+{a}}{{b}},\:\frac{{a}+{b}}{{c}}\:\:{are}\:{roots}\:{of} \\ $$$${cubic}\:{equation}: \\ $$$${q}^{\mathrm{3}} −\mathrm{2008}{q}^{\mathrm{2}} +\mathrm{7056}{q}−\left({xyz}+\mathrm{6056}\right)=\mathrm{0} \\ $$$$….{may}\:\:{continue}.. \\ $$
Commented by jota+ last updated on 03/Dec/17
The end is xyz=(((a+b)(b+c)(c+a))/(abc)) −6056.
$${The}\:{end}\:{is} \\ $$$${xyz}=\frac{\left({a}+{b}\right)\left({b}+{c}\right)\left({c}+{a}\right)}{{abc}}\:−\mathrm{6056}. \\ $$$$ \\ $$$$ \\ $$

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