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Question Number 68912 by TawaTawa last updated on 16/Sep/19

	Answered by MJS last updated on 17/Sep/19
	$(1) a^2 +ab+b^2 −7=0 (2) b^2 +bc+c^2 −21=0 (3) c^2 +ca+a^2 −28=0 (2)−(3) ⇒ c=(7/(a−b))−(a+b) ⇒ (1) a^2 +ab+b^2 −7=0 (2) a^2 +ab+b^2 −35−7((3ab−3b^2 −7)/((a−b)^2 ))=0 (3)=(2) (2)−(1) −28−7((3ab−3b^2 −7)/((a−b)^2 ))=0 b^2 −5ab+4a^2 −7=0 b=((5a)/2)±((√(9a^2 +28))/2) (1) 12a^2 ±3a(√(9a^2 +28))=0 ⇒ a=0∨a=−2∨a=2 ((a),(b),(c) ) ∈{ ((0),((−(√7))),((2(√7))) ) , ((2),(1),(4) ) , (((−2)),((−1)),((−4)) ) , ((0),((√7)),((−2(√7))) )} but a, b, c >0 ⇒ a=2, b=1, c=4 ⇒ n=7 ⇒ n^3 =343$
	$(1) a^{2} + a b + b^{2} - 7 = 0$ $(2) b^{2} + b c + c^{2} - 21 = 0$ $(3) c^{2} + c a + a^{2} - 28 = 0$ $(2) - (3) \Rightarrow c = \frac{7}{a - b} - (a + b)$ $\Rightarrow$ $(1) a^{2} + a b + b^{2} - 7 = 0$ $(2) a^{2} + a b + b^{2} - 35 - 7 \frac{3 a b - 3 b^{2} - 7}{{(a - b)}^{2}} = 0$ $(3) = (2)$ $(2) - (1)$ $- 28 - 7 \frac{3 a b - 3 b^{2} - 7}{{(a - b)}^{2}} = 0$ $b^{2} - 5 a b + 4 a^{2} - 7 = 0$ $b = \frac{5 a}{2} \pm \frac{\sqrt{9 a^{2} + 28}}{2}$ $(1) 12 a^{2} \pm 3 a \sqrt{9 a^{2} + 28} = 0$ $\Rightarrow a = 0 \lor a = - 2 \lor a = 2$ $(\begin{matrix} a \\ b \\ c \end{matrix}) \in {(\begin{matrix} 0 \\ - \sqrt{7} \\ 2 \sqrt{7} \end{matrix}), (\begin{matrix} 2 \\ 1 \\ 4 \end{matrix}), (\begin{matrix} - 2 \\ - 1 \\ - 4 \end{matrix}), (\begin{matrix} 0 \\ \sqrt{7} \\ - 2 \sqrt{7} \end{matrix})}$ $but a, b, c > 0 \Rightarrow$ $a = 2, b = 1, c = 4$ $\Rightarrow n = 7 \Rightarrow n^{3} = 343$
	Commented by TawaTawa last updated on 17/Sep/19

	$Wow, God bless you sir$
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