+26367 Simplify
\(\sqrt[3]{9 + 4 \sqrt{5}}\)
\(\text{Set $X = \sqrt[3]{9 + 4 \sqrt{5}} $} \\ \text{Set $Y = \sqrt[3]{9 - 4 \sqrt{5}} $}\)
Note that \(X\pm Y > 0\)
Now we form \(X+Y\) and \(X-Y\) and consider that
\(\begin{array}{|rcll|} \hline XY &=& \sqrt[3]{9 + 4 \sqrt{5}}\sqrt[3]{9 - 4 \sqrt{5}} \\ XY &=& \sqrt[3]{(9 + 4 \sqrt{5})(9 - 4 \sqrt{5})} \\ XY &=& \sqrt[3]{81-16\cdot 5} \\ XY &=& \sqrt[3]{1} \\ \mathbf{XY} &=& \mathbf{1} \\ \hline \end{array}\\ \begin{array}{|rcll|} \hline X^3+Y^3 &=& (9 + 4 \sqrt{5})+(9 - 4 \sqrt{5}) \\ \mathbf{X^3+Y^3} &=& \mathbf{18} \\ \hline \end{array} \begin{array}{|rcll|} \hline X^3-Y^3 &=& (9 + 4 \sqrt{5})-(9 - 4 \sqrt{5}) \\ \mathbf{X^3-Y^3} &=& \mathbf{8\sqrt{5}} \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline (X+Y)^3 &=& X^3+3X^2Y+3XY^2+Y^3 \\ (X+Y)^3 &=& X^3+Y^3+3X(XY)+3Y(XY) \quad | \quad \mathbf{XY=1} \\ (X+Y)^3 &=& X^3+Y^3+3X +3Y \quad | \quad \mathbf{X^3+Y^3=18} \\ (X+Y)^3 &=& 18+3(X+Y) \\ (X+Y)^3 -3(X+Y) -18&=& 0 \\ \Rightarrow \mathbf{X+Y = 3} && 3^3-3\cdot 3 - 18 = 0\ \checkmark \\ \hline \end{array} \)
\(\begin{array}{|rcll|} \hline (X-Y)^3 &=& X^3-3X^2Y+3XY^2-Y^3 \\ (X-Y)^3 &=& X^3-Y^3-3X(XY)+3Y(XY) \quad | \quad \mathbf{XY=1} \\ (X-Y)^3 &=& X^3-Y^3-3X +3Y \quad | \quad \mathbf{X^3-Y^3=8\sqrt{5}} \\ (X-Y)^3 &=&8\sqrt{5}-3(X-Y) \\ (X-Y)^3+3(X-Y)-8\sqrt{5}&=& 0 \\ \Rightarrow \mathbf{X-Y = \sqrt{5}} && (\sqrt{5})^3+3\sqrt{5} -8\sqrt{5} = 0 \\ && 5\sqrt{5}+3\sqrt{5} -8\sqrt{5} = 0\ \checkmark \\ \hline \end{array}\)
\(\begin{array}{|lrcll|} \hline & \mathbf{X+Y} &=& \mathbf{3} & (1) \\ & \mathbf{X-Y} &=& \mathbf{\sqrt{5} } & (2) \\ \hline (1)+(2): & 2X &=& 3+\sqrt{5} \\ & \mathbf{X} &=& \mathbf{\dfrac{3+\sqrt{5}}{2} } \\ \hline (1)-(2): & 2Y &=& 3-\sqrt{5} \\ & \mathbf{Y} &=& \mathbf{\dfrac{3-\sqrt{5}}{2} } \\ \hline \end{array}\)
\(\begin{array}{l} \sqrt[3]{9 + 4 \sqrt{5}} = \mathbf{\dfrac{3+\sqrt{5}}{2} } \\\\ \sqrt[3]{9 - 4 \sqrt{5}} = \mathbf{\dfrac{3-\sqrt{5}}{2} } \end{array}\)
