{ ((a^3 =3ab^2 +11)),((b^3 =3a^2 b+2)) :} ; a,b ∈R ⇒a^2 +b^2 =?


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Question Number 176988 by cortano1 last updated on 29/Sep/22
${ ((a^3 =3ab^2 +11)),((b^3 =3a^2 b+2)) :} ; a,b ∈R ⇒a^2 +b^2 =?$
${\begin{cases} a^{3} = 3 {a b}^{2} + 11 \\ b^{3} = 3 a^{2} b + 2 \end{cases}; a, b \in R$ $\Rightarrow a^{2} + b^{2} = ?$
Commented by mr W last updated on 29/Sep/22

$w r o n g!$ $a^{3} - 3 {a b}^{2} = 11 \Rightarrow a (a^{2} - 3 b^{2}) = 11 (1)$ $b^{3} - 3 a^{2} b = 2 \Rightarrow - b (3 a^{2} - b^{2}) = 2$
Commented by a.lgnaoui last updated on 29/Sep/22

$a^{3} - 3 {a b}^{2} = 11 \Rightarrow a (a^{2} - b^{2}) = 11 (1)$ $b^{3} - 3 a^{2} b = 2 \Rightarrow - b (a^{2} - b^{2}) = 2$ $\frac{a}{11} + \frac{b}{2} = 0 b = - \frac{2}{11} a \Rightarrow a^{2} + b^{2} = \frac{125}{121} a^{2} (2)$ $(1) a^{2} - b^{2} = \frac{11}{a}$ $(2) a^{2} + b^{2} = \frac{125}{121} a^{2}$ $2 a^{2} = \frac{11}{a} + \frac{125}{11^{2}} a^{2} 2 \times 11^{2} a^{3} - 125 a^{3} - 11^{3} = 0$ $117 a^{3} - 11^{3} = 0 \Rightarrow a = \frac{11}{^{3} \sqrt{117}}$ $(2) \Leftrightarrow a^{2} + b^{2} = \frac{125}{11^{2}} \times {(\frac{11}{^{3} \sqrt{117}})}^{2} = \frac{125}{23, 91}$ $a^{2} + b^{2} = 5, 227$
	Answered by blackmamba last updated on 29/Sep/22
	$⇒ { ((a^3 −3ab^2 =11)),((b^3 −3a^2 b=2)) :} ⇒(a^2 +b^2 )^3 =(a^2 −3ab^2 )^2 +(b^3 −3a^2 b)^2 ⇒a^2 +b^2 =((11^2 +2^2 ))^(1/3) =((125))^(1/3) = 5$
	$\Rightarrow {\begin{cases} a^{3} - 3 {a b}^{2} = 11 \\ b^{3} - 3 a^{2} b = 2 \end{cases}$ $\Rightarrow {(a^{2} + b^{2})}^{3} = {(a^{2} - 3 {a b}^{2})}^{2} + {(b^{3} - 3 a^{2} b)}^{2}$ $\Rightarrow a^{2} + b^{2} = \sqrt[3]{11^{2} + 2^{2}} = \sqrt[3]{125} = 5$
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