Question-173629

Question Number 173629 by mnjuly1970 last updated on 15/Jul/22

Commented by mnjuly1970 last updated on 15/Jul/22

s o l v e f o r R a, b, c ? ⇑⇑⇑

Commented by Tawa11 last updated on 15/Jul/22

Great sirs

Answered by mahdipoor last updated on 15/Jul/22

a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b) =

a + b + (c) = a + b + (a + b) = 2 (a + b)

i \Rightarrow a + b = 0 \Rightarrow a = k, b = - k, c = 0

(a^{2} + b^{2} = b + c) \Rightarrow 2 k^{2} = - k \Rightarrow k = 0 o r - \frac{1}{2}

\Rightarrow\Rightarrow (\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) o r (\begin{matrix} - 0.5 \\ 0.5 \\ 0 \end{matrix})

ii \Rightarrow a + b \neq 0 \Rightarrow 2 = (a^{2} + b^{2}) - a b = b + (c) - a b

= 2 b + a - a b \Rightarrow 2 b + a - a b - 2 = 0 \Rightarrow

(a - 2) (1 - b) = 0 \Rightarrow

1) i f a = 2 \Rightarrow c = 2 + b & 4 + b^{2} = b + c \Rightarrow

4 + b^{2} = 2 b + 2 \Rightarrow b^{2} - 2 b + 2 = 0 \Rightarrow ∄ b \in R

2) i f b = 1 \Rightarrow c = 1 + a & 1 + a^{2} = 1 + c \Rightarrow

1 + a^{2} = 2 + a \Rightarrow a^{2} - a - 1 = 0 \Rightarrow a = \frac{1 \pm \sqrt{5}}{2}

\Rightarrow\Rightarrow (\begin{matrix} a \\ b \\ c \end{matrix}) = (\begin{matrix} \frac{1 \pm \sqrt{5}}{2} \\ 1 \\ \frac{3 \pm \sqrt{5}}{2} \end{matrix})

Answered by MJS_new last updated on 15/Jul/22

obviously a = b = c = 0

c = a + b

b = p a

{\begin{cases} (p^{2} + 1) a^{2} - (2 p + 1) a = 0 \\ (p^{3} + 1) a^{3} - 2 (p + 1) a = 0 \end{cases}

a \neq 0

{\begin{cases} (p^{2} + 1) a - 2 p - 1 = 0 \\ (p + 1) ((p^{2} - p + 1) a^{2} - 2) = 0 \end{cases}

p_{1} = - 1 \Rightarrow a_{1} = - \frac{1}{2} \land b_{1} = \frac{1}{2} \land c_{1} = 0

a = \frac{2 p + 1}{p^{2} + 1}

p^{4} - \frac{3}{2} p^{2} + \frac{3}{2} p - \frac{1}{2} = 0

(p^{2} + p - 1) (p^{2} - p + \frac{1}{2}) = 0

p_{2} = - \frac{1}{2} - \frac{\sqrt{5}}{2} \Rightarrow a_{2} = \frac{1}{2} - \frac{\sqrt{5}}{2} \land b_{2} = 1 \land c_{2} = \frac{3}{2} - \frac{\sqrt{5}}{2}

p_{3} = - \frac{1}{2} + \frac{\sqrt{5}}{2} \Rightarrow a_{3} = \frac{1}{2} + \frac{\sqrt{5}}{2} \land b_{3} = 1 \land c_{3} = \frac{3}{2} + \frac{\sqrt{5}}{2}

(p_{4, 5} = \frac{1}{2} \pm \frac{1}{2} i)

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