Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
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
	$obviously a=b=c=0 c=a+b b=pa { (((p^2 +1)a^2 −(2p+1)a=0)),(((p^3 +1)a^3 −2(p+1)a=0)) :} a≠0 { (((p^2 +1)a−2p−1=0)),(((p+1)((p^2 −p+1)a^2 −2)=0)) :} p_1 =−1 ⇒ a_1 =−(1/2)∧b_1 =(1/2)∧c_1 =0 a=((2p+1)/(p^2 +1)) p^4 −(3/2)p^2 +(3/2)p−(1/2)=0 (p^2 +p−1)(p^2 −p+(1/2))=0 p_2 =−(1/2)−((√5)/2) ⇒ a_2 =(1/2)−((√5)/2)∧b_2 =1∧c_2 =(3/2)−((√5)/2) p_3 =−(1/2)+((√5)/2) ⇒ a_3 =(1/2)+((√5)/2)∧b_3 =1∧c_3 =(3/2)+((√5)/2) (p_(4, 5) =(1/2)±(1/2)i)$
	$obviously a = b = c = 0$ $c = a + b$ $b = p a$ ${\begin{cases} (p^{2} + 1) a^{2} - (2 p + 1) a = 0 \\ (p^{3} + 1) a^{3} - 2 (p + 1) a = 0 \end{cases}$ $a \neq 0$ ${\begin{cases} (p^{2} + 1) a - 2 p - 1 = 0 \\ (p + 1) ((p^{2} - p + 1) a^{2} - 2) = 0 \end{cases}$ $p_{1} = - 1 \Rightarrow a_{1} = - \frac{1}{2} \land b_{1} = \frac{1}{2} \land c_{1} = 0$ $a = \frac{2 p + 1}{p^{2} + 1}$ $p^{4} - \frac{3}{2} p^{2} + \frac{3}{2} p - \frac{1}{2} = 0$ $(p^{2} + p - 1) (p^{2} - p + \frac{1}{2}) = 0$ $p_{2} = - \frac{1}{2} - \frac{\sqrt{5}}{2} \Rightarrow a_{2} = \frac{1}{2} - \frac{\sqrt{5}}{2} \land b_{2} = 1 \land c_{2} = \frac{3}{2} - \frac{\sqrt{5}}{2}$ $p_{3} = - \frac{1}{2} + \frac{\sqrt{5}}{2} \Rightarrow a_{3} = \frac{1}{2} + \frac{\sqrt{5}}{2} \land b_{3} = 1 \land c_{3} = \frac{3}{2} + \frac{\sqrt{5}}{2}$ $(p_{4, 5} = \frac{1}{2} \pm \frac{1}{2} i)$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum