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Question-153153




Question Number 153153 by liberty last updated on 05/Sep/21
Answered by mr W last updated on 05/Sep/21
(1)  a+b+c=4  (a+b+c)^2 =a^2 +b^2 +c^2 +2(ab+bc+ca)  4^2 =14+2(ab+bc+ca)  ⇒ab+bc+ca=1  (a+b+c)^3 =a^3 +b^3 +c^3 +3(a^2 b+ab^2 +b^2 c+bc^2 +c^2 a+ca^2 )+6abc  (a+b+c)^3 =a^3 +b^3 +c^3 +3(ab+bc+ca)(a+b+c)−3abc  4^3 =34+3×1×4−3abc  ⇒abc=−6  a,b,c are roots of  x^3 −4x^2 +x+6=0  (x+1)(x−2)(x−3)=0  x=−1,2,3  ⇒(a,b,c)=(−1,2,3)=(−1,3,2)=(2,−1,3)=(2,3,−1)=(3,−1,2)=(3,2,−1)
$$\left(\mathrm{1}\right) \\ $$$${a}+{b}+{c}=\mathrm{4} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\mathrm{4}^{\mathrm{2}} =\mathrm{14}+\mathrm{2}\left({ab}+{bc}+{ca}\right) \\ $$$$\Rightarrow{ab}+{bc}+{ca}=\mathrm{1} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({a}^{\mathrm{2}} {b}+{ab}^{\mathrm{2}} +{b}^{\mathrm{2}} {c}+{bc}^{\mathrm{2}} +{c}^{\mathrm{2}} {a}+{ca}^{\mathrm{2}} \right)+\mathrm{6}{abc} \\ $$$$\left({a}+{b}+{c}\right)^{\mathrm{3}} ={a}^{\mathrm{3}} +{b}^{\mathrm{3}} +{c}^{\mathrm{3}} +\mathrm{3}\left({ab}+{bc}+{ca}\right)\left({a}+{b}+{c}\right)−\mathrm{3}{abc} \\ $$$$\mathrm{4}^{\mathrm{3}} =\mathrm{34}+\mathrm{3}×\mathrm{1}×\mathrm{4}−\mathrm{3}{abc} \\ $$$$\Rightarrow{abc}=−\mathrm{6} \\ $$$${a},{b},{c}\:{are}\:{roots}\:{of} \\ $$$${x}^{\mathrm{3}} −\mathrm{4}{x}^{\mathrm{2}} +{x}+\mathrm{6}=\mathrm{0} \\ $$$$\left({x}+\mathrm{1}\right)\left({x}−\mathrm{2}\right)\left({x}−\mathrm{3}\right)=\mathrm{0} \\ $$$${x}=−\mathrm{1},\mathrm{2},\mathrm{3} \\ $$$$\Rightarrow\left({a},{b},{c}\right)=\left(−\mathrm{1},\mathrm{2},\mathrm{3}\right)=\left(−\mathrm{1},\mathrm{3},\mathrm{2}\right)=\left(\mathrm{2},−\mathrm{1},\mathrm{3}\right)=\left(\mathrm{2},\mathrm{3},−\mathrm{1}\right)=\left(\mathrm{3},−\mathrm{1},\mathrm{2}\right)=\left(\mathrm{3},\mathrm{2},−\mathrm{1}\right) \\ $$

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