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

S o l v e f o r r e a l n u m b e r s :

{\begin{cases} - a^{1} - b^{2} - c^{3} = 2024^{12} \\ - a^{- 1} - b^{- 2} - c^{- 3} = 2024^{- 12} \\ a^{1} b^{2} c^{3} = 2024^{24} \end{cases}

Answered by MJS_new last updated on 18/Apr/21

- a - b^{2} - c^{3} = p

- \frac{1}{a} - \frac{1}{b^{2}} - \frac{1}{c^{3}} = \frac{1}{p}

{a b}^{2} c^{3} = p^{2}

easy to solve, I get

a = - \sqrt{p} \land b = \sqrt[4]{p} \land c = - \sqrt[3]{p}

p = 2024^{12}

\Rightarrow

a = - 2024^{6} \land b = 2024^{3} \land c = - 2024^{4}

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