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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)

(1)a+b+c=4(a+b+c)2=a2+b2+c2+2(ab+bc+ca)42=14+2(ab+bc+ca)⇒ab+bc+ca=1(a+b+c)3=a3+b3+c3+3(a2b+ab2+b2c+bc2+c2a+ca2)+6abc(a+b+c)3=a3+b3+c3+3(ab+bc+ca)(a+b+c)−3abc43=34+3×1×4−3abc⇒abc=−6a,b,carerootsofx3−4x2+x+6=0(x+1)(x−2)(x−3)=0x=−1,2,3⇒(a,b,c)=(−1,2,3)=(−1,3,2)=(2,−1,3)=(2,3,−1)=(3,−1,2)=(3,2,−1)

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