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Question Number 109948 by mnjuly1970 last updated on 26/Aug/20

	Answered by 1549442205PVT last updated on 26/Aug/20
	$acosA=a.((b^2 +c^2 −a^2 )/(2bc))=((a^2 (b^2 +c^2 −a^2 ))/(2abc)). Similarly,we have the other equalities for b,c Hence acosA+bcosB+c.cosC=Σ_(cyc) ((a^2 (b^2 +c^2 −a^2 ))/(2abc)) = ((a^2 b^2 +b^2 c^2 +c^2 a^2 −0.5(a^4 +b^4 +c^4 ))/(abc)) (∗) a,b,c are roots of the equation x^3 −24x^2 +180x−420 by Viete′s theorem we have { ((a+b+c=24)),((ab+bc+ca=180)),((abc=420)) :} a^2 +b^2 +c^2 =(a+b+c)^2 −2(ab+bc+ca) =24^2 −2.180=576−360=216 (a^2 b^2 +b^2 c^2 +c^2 a^2 )=(ab+bc+ca)^2 −2abc(a+b+c) =180^2 −2.420.24=32400−20160=12240(1) (a^4 +b^4 +c^4 )=(a^2 +b^2 +c^2 )^2 −2(a^2 b^2 +b^2 c^2 +c^2 a^2 ) =216^2 −2.12240=22176(2) Replace (1)(2)and abc =420 into (∗) we get ((a^2 b^2 +b^2 c^2 +c^2 a^2 −(a^4 +b^4 +c^4 ))/(abc)) =((12240−0.5×22176)/(420))=((96)/(35)) Thus,acosA+bcosB+c.cosC=((96)/(35))$
	$acosA = a . \frac{b^{2} + c^{2} - a^{2}}{2 bc} = \frac{a^{2} (b^{2} + c^{2} - a^{2})}{2 abc} .$ $Similarly, we have the other equalities$ $for b, c Hence$ $acosA + bcosB + c . cosC = \underset{cyc}{Σ} \frac{a^{2} (b^{2} + c^{2} - a^{2})}{2 abc}$ $= \frac{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} - 0.5 (a^{4} + b^{4} + c^{4})}{abc} ()$ $a, b, c are roots of the equation$ $x^{3} - {24 x}^{2} + 180 x - 420 by {Viete}^{'} s theorem$ $we have {\begin{cases} a + b + c = 24 \\ ab + bc + ca = 180 \\ abc = 420 \end{cases}$ $a^{2} + b^{2} + c^{2} = {(a + b + c)}^{2} - 2 (ab + bc + ca)$ $= 24^{2} - 2.180 = 576 - 360 = 216$ $(a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2}) = {(ab + bc + ca)}^{2} - 2 abc (a + b + c)$ $= 180^{2} - 2.420.24 = 32400 - 20160 = 12240 (1)$ $(a^{4} + b^{4} + c^{4}) = {(a^{2} + b^{2} + c^{2})}^{2} - 2 (a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2})$ $= 216^{2} - 2.12240 = 22176 (2)$ $Replace (1) (2) and abc = 420 into ()$ $we get \frac{a^{2} b^{2} + b^{2} c^{2} + c^{2} a^{2} - (a^{4} + b^{4} + c^{4})}{abc}$ $= \frac{12240 - 0.5 \times 22176}{420} = \frac{96}{35}$ $Thus, acosA + bcosB + c . cosC = \frac{96}{35}$
	Commented by mnjuly1970 last updated on 27/Aug/20

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