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Question Number 106943 by hgrocks last updated on 08/Aug/20

Commented by hgrocks last updated on 08/Aug/20

	Answered by 1549442205PVT last updated on 09/Aug/20
	$Set a=α,b=β,c=γ.From the hypothesis we have: { ((a+b+c=5 (1))),((a^2 +b^2 +c^2 =19 (2))),((S=a^3 +b^3 +c^3 )) :} We have :ab+bc+ca= [(a+b+c)^2 −(a^2 +b^2 +c^2 )]/2=3 (3) S=a^3 +b^3 +c^3 =(a+b)^3 −3ab(a+b) +[5−(a+b)]^3 (4) From(1) and (3)we get ab+bc+ca=3 ⇔ab+(a+b)[5−(a+b)]=3 ⇒ab=(a+b)^2 −5(a+b)+3 Putting a+b=t we have: ab=t^2 −5t+3 (5).Therefore,from(4) we get S=t^3 −3t(t^2 −5t+3)+(5−t)^3 =t^3 −3t(t^2 −5t+3)+125−75t+15t^2 −t^3 =−3t^3 +30t^2 −84t+125 =3(5−t)[t^2 −5t+3)+80 =3(5−t)[(5−t)^2 −5(5−t)+3]+80 =3c(c^2 −5c+3)+80=3(c^3 −5c^2 +3c)+80(∗) Since three numbers are equal in role WLOG we can suppose that c≥0.Also, From (5)we have ab=(5−c)^2 −5(5−c)+3 =c^2 −5c+3 ⇒(a−b)^2 =(a+b)^2 −4ab =(5−c)^2 −4(c^2 −5c+3)=−3c^2 +10c+13≥0 ⇔−3(c−(5/3))^2 +((64)/3)≥0⇔(3c−5)^2 ≤64 ⇔∣3c−5∣≤8⇒3c−13≤0⇒0≤c≤((13)/3) (6) Set f(c)=c^3 −5c^2 +3c with c∈[0,((13)/3)]we have f ′(c)=3c^2 −10c+3 =0⇔c∈{(1/3);3} Then we have the variable table as: determinant ((x,0,,(1/3),,3,((13)/3)),((f ′(c)),3,(+↗),0,(−↘),↗,),((f(c)),0,,((13)/(27)),,(−9),((13)/(27)))) From tablet we see that f(c)_(max) =((13)/(27)) when c=(1/3)or c=((13)/(27)) Therefore,from (∗)we get S≤80+3.((13)/(27))=80((13)/9) Thus,S_(max) =80 ((13)/9) i)when c=(1/3),a+b=((14)/3) Combining to (5) we get ab=((13)/9) ⇒(a−b)^2 =(a+b)^2 −4ab=((196)/9)−4×((13)/9)=((144)/9) ⇒a−b=((12 )/3) (suppose a≥b) ⇒a=((13)/3),b=c=(1/3) ii)When c=((13)/3)⇒a+b=(2/3)⇒a=b=(1/3) ⇒S_(max) =80 if and only if (a,b,c)=(((13)/3),(1/3),(1/3))and all permutations of them Other way: S=3(c^3 −5c^2 +3c)+80(∗) =(1/9)(27c^3 −135c^2 +81c)+80 =(1/9){[(3c−1)^2 (3c−13)]+13}+80 =(1/9)[(3c−1)^2 (3c−13)]+80((13)/9) From the condition (6) we infer S=(1/9)[(3c−1)^2 (3c−13)]+80((13)/9) ≤80((13)/9) since (3c−1)^2 (3c−13)≤0 due to 3c≤13$
	$Set a = α, b = β, c = γ . From the$ $hypothesis we have : {\begin{cases} a + b + c = 5 (1) \\ a^{2} + b^{2} + c^{2} = 19 (2) \\ S = a^{3} + b^{3} + c^{3} \end{cases}$ $We have : ab + bc + ca =$ $[{(a + b + c)}^{2} - (a^{2} + b^{2} + c^{2})] / 2 = 3 (3)$ $S = a^{3} + b^{3} + c^{3} = {(a + b)}^{3} - 3 ab (a + b)$ $+ {[5 - (a + b)]}^{3} (4)$ $From (1) and (3) we get ab + bc + ca = 3$ $\Leftrightarrow ab + (a + b) [5 - (a + b)] = 3$ $\Rightarrow ab = {(a + b)}^{2} - 5 (a + b) + 3$ $Putting a + b = t we have :$ $ab = t^{2} - 5 t + 3 (5) . Therefore, from (4)$ $we get S = t^{3} - 3 t (t^{2} - 5 t + 3) + {(5 - t)}^{3}$ $= t^{3} - 3 t (t^{2} - 5 t + 3) + 125 - 75 t + {15 t}^{2} - t^{3}$ $= - {3 t}^{3} + {30 t}^{2} - 84 t + 125$ $= 3 (5 - t) [t^{2} - 5 t + 3) + 80$ $= 3 (5 - t) [{(5 - t)}^{2} - 5 (5 - t) + 3] + 80$ $= 3 c (c^{2} - 5 c + 3) + 80 = 3 (c^{3} - {5 c}^{2} + 3 c) + 80 ()$ $Since three numbers are equal in role$ $WLOG we can suppose that c ⩾ 0 . Also,$ $From (5) we have ab = {(5 - c)}^{2} - 5 (5 - c) + 3$ $= c^{2} - 5 c + 3 \Rightarrow {(a - b)}^{2} = {(a + b)}^{2} - 4 ab$ $= {(5 - c)}^{2} - 4 (c^{2} - 5 c + 3) = - {3 c}^{2} + 10 c + 13 ⩾ 0$ $\Leftrightarrow - 3 {(c - \frac{5}{3})}^{2} + \frac{64}{3} ⩾ 0 \Leftrightarrow {(3 c - 5)}^{2} ⩽ 64$ $\Leftrightarrow∣ 3 c - 5 ∣⩽ 8 \Rightarrow 3 c - 13 ⩽ 0 \Rightarrow 0 ⩽ c ⩽ \frac{13}{3} (6)$ $Set f (c) = c^{3} - {5 c}^{2} + 3 c with c \in [0, \frac{13}{3}] we have$ $f^{'} (c) = {3 c}^{2} - 10 c + 3 = 0 \Leftrightarrow c \in {\frac{1}{3}; 3}$ $Then we have the variable table as :$ $\| \begin{matrix} x & 0 & \frac{1}{3} & 3 & \frac{13}{3} \\ f^{'} (c) & 3 & + ↗ & 0 & - ↘ & ↗ \\ f (c) & 0 & \frac{13}{27} & - 9 & \frac{13}{27} \end{matrix} \|$ $From tablet we see that f {(c)}_{\max} = \frac{13}{27}$ $when c = \frac{1}{3} or c = \frac{13}{27}$ $Therefore, from () we get S ⩽ 80 + 3 . \frac{13}{27} = 80 \frac{13}{9}$ $Thus, S_{\max} = 80 \frac{13}{9}$ $i) when c = \frac{1}{3}, a + b = \frac{14}{3}$ $Combining to (5) we get ab = \frac{13}{9}$ $\Rightarrow {(a - b)}^{2} = {(a + b)}^{2} - 4 ab = \frac{196}{9} - 4 \times \frac{13}{9} = \frac{144}{9}$ $\Rightarrow a - b = \frac{12}{3} (suppose a ⩾ b)$ $\Rightarrow a = \frac{13}{3}, b = c = \frac{1}{3}$ $ii) When c = \frac{13}{3} \Rightarrow a + b = \frac{2}{3} \Rightarrow a = b = \frac{1}{3}$ $\Rightarrow S_{\max} = 80 if and only if$ $(a, b, c) = (\frac{13}{3}, \frac{1}{3}, \frac{1}{3}) and all$ $permutations of them$ $Other way :$ $S = 3 (c^{3} - {5 c}^{2} + 3 c) + 80 (*)$ $= \frac{1}{9} ({27 c}^{3} - {135 c}^{2} + 81 c) + 80$ $= \frac{1}{9} {[{(3 c - 1)}^{2} (3 c - 13)] + 13} + 80$ $= \frac{1}{9} [{(3 c - 1)}^{2} (3 c - 13)] + 80 \frac{13}{9}$ $From the condition (6) we infer$ $S = \frac{1}{9} [{(3 c - 1)}^{2} (3 c - 13)] + 80 \frac{13}{9} ⩽ 80 \frac{13}{9}$ $since {(3 c - 1)}^{2} (3 c - 13) ⩽ 0 due to 3 c ⩽ 13$
	Answered by mr W last updated on 08/Aug/20

	$l e t p_{n} = α^{n} + β^{n} + γ^{n}$ $p_{1} = e_{1} = 5$ $p_{2} = 19 = e_{1} p_{1} - 2 e_{2} \Rightarrow e_{2} = \frac{p_{1}^{2} - p_{2}}{2}$ $p_{3} = e_{1} p_{2} - e_{2} p_{1} + 3 e_{3}$ $= p_{1} p_{2} - \frac{(p_{1}^{2} - p_{2}) p_{1}}{2} + 3 e_{3}$ $\Rightarrow p_{3} = \frac{p_{1} (3 p_{2} - p_{1}^{2})}{2} + 3 e_{3} = 80 + 3 e_{3}$ $w i t h e_{3} = α β γ$ $γ = 5 - α - β$ $e_{3, m a x / m i n} h a p p e n s a t α = β$ $2 α^{2} + {(5 - 2 α)}^{2} = 19$ $3 α^{2} - 10 α + 3 = 0$ $\Rightarrow α = \frac{5 \pm 4}{3} = 3, \frac{1}{3} \Rightarrow γ = - 1, \frac{13}{3}$ $e_{3, m i n} = 3^{2} \times (- 1) = - 9$ $e_{3, m a x} = {(\frac{1}{3})}^{2} \times (\frac{13}{3}) = \frac{13}{27}$ $\Rightarrow p_{3, m i n} = 80 - 27 = 53$ $\Rightarrow p_{3, m a x} = 80 + \frac{13}{9} = \frac{733}{9}$
	Answered by mr W last updated on 08/Aug/20

	$M e t h o d 2$ $γ = 5 - α - β$ $α^{2} + β^{2} + {(5 - α - β)}^{2} - 19 = 0$ $S = α^{3} + β^{3} + {(5 - α - β)}^{2}$ $S_{m a x} i s a t α = β$ $\Rightarrow 2 α^{2} + {(5 - 2 α)}^{2} - 19 = 0$ $\Rightarrow α = β = 3, \frac{1}{3}$ $\Rightarrow γ = - 1, \frac{13}{3}$ $\Rightarrow S_{m i n} = 2 \times 3^{3} + {(- 1)}^{3} = 53$ $\Rightarrow S_{m a x} = 2 \times {(\frac{1}{3})}^{3} + {(\frac{13}{3})}^{3} = \frac{733}{9}$
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