Question

Cum s-ar rationaliza in situatia data? (ordin 3)​

Answer 1

$\it a^3-b^3=(a-b)(a^2+ab+b^2)\\ \\ ((\sqrt[3]4)^3-\sqrt[3]2)=(\sqrt[3]4-\sqrt[3]2)(\sqrt[3]{16}+\sqrt[3]8+\sqrt[3]4)\ \ \ \ \ (1)\\ \\ Dar,\ (\sqrt[3]4)^3-(\sqrt[3]2)^3=4-2=2\ \ \ \ \ (2)\\ \\ (1),(2) \Rightarrow (\sqrt[3]4-\sqrt[3]2) =2$

Prin urmare, vom amplifica fracția cu (∛16 + ∛8 + ∛4) și rezultă:

$\dfrac{4(\sqrt[3]{16}+\sqrt[3]8+\sqrt[3]4)}{2} =2(\sqrt[3]{16}+\sqrt[3]8+\sqrt[3]4)$