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Solve-for-real-numbers-a-1-b-2-c-3-2024-12-a-1-b-2-c-3-2024-12-a-1-b-2-c-3-2024-24-




Question Number 138827 by mathdanisur last updated on 18/Apr/21
Solve for real numbers:   { ((−a^1 −b^2 −c^3 =2024^(12) )),((−a^(−1) −b^(−2) −c^(−3) =2024^(−12) )),((a^1 b^2 c^3 =2024^(24) )) :}
$${Solve}\:{for}\:{real}\:{numbers}: \\ $$$$\begin{cases}{−{a}^{\mathrm{1}} −{b}^{\mathrm{2}} −{c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{12}} }\\{−{a}^{−\mathrm{1}} −{b}^{−\mathrm{2}} −{c}^{−\mathrm{3}} =\mathrm{2024}^{−\mathrm{12}} }\\{{a}^{\mathrm{1}} {b}^{\mathrm{2}} {c}^{\mathrm{3}} =\mathrm{2024}^{\mathrm{24}} }\end{cases} \\ $$
Answered by MJS_new last updated on 18/Apr/21
−a−b^2 −c^3 =p  −(1/a)−(1/b^2 )−(1/c^3 )=(1/p)  ab^2 c^3 =p^2   easy to solve, I get  a=−(√p)∧b=(p)^(1/4) ∧c=−(p)^(1/3)   p=2024^(12)   ⇒  a=−2024^6 ∧b=2024^3 ∧c=−2024^4
$$−{a}−{b}^{\mathrm{2}} −{c}^{\mathrm{3}} ={p} \\ $$$$−\frac{\mathrm{1}}{{a}}−\frac{\mathrm{1}}{{b}^{\mathrm{2}} }−\frac{\mathrm{1}}{{c}^{\mathrm{3}} }=\frac{\mathrm{1}}{{p}} \\ $$$${ab}^{\mathrm{2}} {c}^{\mathrm{3}} ={p}^{\mathrm{2}} \\ $$$$\mathrm{easy}\:\mathrm{to}\:\mathrm{solve},\:\mathrm{I}\:\mathrm{get} \\ $$$${a}=−\sqrt{{p}}\wedge{b}=\sqrt[{\mathrm{4}}]{{p}}\wedge{c}=−\sqrt[{\mathrm{3}}]{{p}} \\ $$$${p}=\mathrm{2024}^{\mathrm{12}} \\ $$$$\Rightarrow \\ $$$${a}=−\mathrm{2024}^{\mathrm{6}} \wedge{b}=\mathrm{2024}^{\mathrm{3}} \wedge{c}=−\mathrm{2024}^{\mathrm{4}} \\ $$

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