Question

Aratati ca ∛(16+8√5) - ∛(16-8√5) = 2√5 .
Ajutor , va rog.

Answer 1

$\sqrt[3]{16+8\sqrt5}-\sqrt[3]{16-8\sqrt5} = 2\sqrt5 \\ \\ Notam a = \sqrt[3]{16+8\sqrt5};\quad b = \sqrt[3]{16-8\sqrt5}\\ \\ a^3-b^3 = 16+8\sqrt5 - (16-8\sqrt5) \\ a^3-b^3 = 16+8\sqrt5-16+8\sqrt5 \\ a^3-b^3 = 16\sqrt5 \\ \\ (a-b)(a^2+ab+b^2) = 16\sqrt5 \\ \\ (a-b)(a^2-2ab+b^2+3ab) = 16\sqrt5 \\ \\ (a-b)\cdot \Big[(a-b)^2+3ab\Big] = 16\sqrt5 \\ \\ (a-b)\cdot \Big[(a-b)^2+3\cdot \sqrt[3]{16+8\sqrt5}\cdot \sqrt[3]{16-8\sqrt5}\Big] = 16\sqrt5 \\ \\ Notam (a-b) = t;$

$t\cdot \Big[t^2+3\cdot \sqrt[3]{(16+8\sqrt5)\cdot (16-8\sqrt5)}\Big] = 16\sqrt5 \\ \\ t\cdot \Big[t^2+3\cdot \sqrt[3]{16^2-(8\sqrt5)^2}\Big] = 16\sqrt5 \\ \\ t\cdot \Big(t^2+3\cdot \sqrt[3]{256 - 320}\Big) = 16\sqrt5\\ \\ t\cdot \Big(t^2+ 3\cdot \sqrt[3]{-64}\Big) = 16\sqrt5 \\ \\ t\cdot \Big[t^2+3\cdot (-4)\Big]=16\sqrt5 \\ \\ t\cdot \Big(t^2-12\Big) = 16\sqrt5$

$\\ t = 2\sqrt5 este relatie adevarata daca t = 2\sqrt5 este solutie\\ pentru ecuatia de mai sus, adica, verifica ecuatia de mai sus. \\ \\t = 2\sqrt5: \\ \\ (2\sqrt5)\cdot \Big[(2\sqrt5)^2-12\Big] = 16\sqrt5\\ \\ (2\sqrt5)\cdot (4\cdot 5-12)=16\sqrt5 \\ \\ (2\sqrt5)\cdot (20-12) = 16\sqrt5 \\ \\ (2\sqrt5)\cdot 8 = 16\sqrt5 \\ \\ 16\sqrt5 = 16\sqrt5\quad (A)\\ \\ \Rightarrow t = 2\sqrt5 \text{(A)} \Leftrightarrow a-b = 2\sqrt5 \text{(A)} \Leftrightarrow \\ \\ \Leftrightarrow \boxed{\sqrt[3]{16+8\sqrt5}-\sqrt[3]{16-8\sqrt5} = 2\sqrt5\quad \text{(A)}}$