An algebra problem by Naman Choudhary

Let $\color{#D61F06}{A=\sqrt[3]{9+4\sqrt5}}$ , $\color{#3D99F6}{B=\sqrt[3]{9-4\sqrt5}}$ and $\color{#20A900}S=\color{#D61F06}A+\color{#3D99F6}B=\color{#D61F06}{\sqrt[3]{9+4\sqrt5}}+\color{#3D99F6}{\sqrt[3]{9-4\sqrt5}}$

Now,

$\begin{aligned}\color{#20A900}S^3=&\ (A+B)^3\\=&\ A^3+3A^2B+3AB^2+B^3\\=&\ A^3+B^3+3AB(\color{#D61F06}A+\color{#3D99F6}B)\\=&\ \color{#302B94}{A^3+B^3}+\color{magenta}{3AB}\color{#20A900}S\Rightarrow(*)\end{aligned}$

Thus,

$\begin{aligned}\color{#302B94}{A^3+B^3}=&\ \left(\color{#D61F06}{\sqrt[3]{9+4\sqrt5}}\right)^3+\left(\color{#3D99F6}{\sqrt[3]{9-4\sqrt5}}\right)^3\\=&\ 9+4\sqrt5+9-4\sqrt5\\=&\ \color{#302B94}{18}\end{aligned}$

$\begin{aligned}\color{magenta}{3AB}=&\ 3\left(\color{#D61F06}{\sqrt[3]{9+4\sqrt5}}\right)\left(\color{#3D99F6}{\sqrt[3]{9-4\sqrt5}}\right)\\=&\ 3\sqrt[3]{(9+4\sqrt5)(9-4\sqrt5)}\\=&\ 3\sqrt[3]{81-80}\\=&\ 3\sqrt[3]1\\=&\ \color{magenta}3\end{aligned}$

Instead $\color{#302B94}{A^3+B^3}$ with $\color{#302B94}{18}$ and $\color{magenta}{3AB}$ with $\color{magenta}3$ in equation $(*)$

$\begin{aligned}\color{#20A900}S^3=&\ \color{#302B94}{18}+\color{magenta}3\color{#20A900}S\\\color{#20A900}S^3-3\color{#20A900}S-18=&\ 0\\(\color{#20A900}S-3)(\color{#20A900}S^2+3\color{#20A900}S+6)=&\ 0\\\color{#20A900}{S=}&\ \color{#20A900}3\end{aligned}$

From equation

$\begin{aligned}\color{#20A900}S^2+3\color{#20A900}S+6=&\ 0\\\Delta=&\ 3^2-4(1)(6)=&-15\end{aligned}$

$\Delta<0$ . Thus, this equation has no real solution.

Hence, $\sqrt[3]{9+4\sqrt5}+\sqrt[3]{9-4\sqrt5}=\boxed3$

set it equal to z, cube the whole thing Some Fundemental polynomials that we need to know 1. (a+b)^3 = a^3 +3a^2 b + 3a b^2 + 3b^3 2. (a+b)(a-b) = a^2 -b^2 If we substitute a for cuberoot(9+4sqrt(5)) and b for cuberoot(9-4 sqrt(5)), z^3 expanded, we get (9+4sqrt(5)) + 3 cuberoot(9+4sqrt(5))^2 * cuberoot(9-4sqrt(5)) + 3*cuberoot(9-4sqrt(5))^2 * cuberoot(9+4sqrt(5)) + (9-4sqrt(5)

now we use the second expansion formula (9+4sqrt(5)) (9-4sqrt(5)) = 81-81 =1 so z^3 = (9+4sqrt(5)) + 3 1 cuberoot(9+4sqrt(5)) + 3 1*cuberoot(9-4sqrt(5)) + 9-4sqrt(5) if we simplify, we get z^3 = 18 + 3cuberoot(9+4sqrt(5)) + 3(9-4sqrt(5))

now we divide everything by 3, z^3 / 3 = 6+cuberoot(9+4sqrt(5)) + (9-4sqrt(5)) that second part, cuberoot(9+4sqrt(5)) + (9-4sqrt(5)), is what we set to z so the difference between z^3 / 3 and z = 6 z^3 /3 - z = 6 (z^3 - 3z)/3 = 6 z^3 -3z-18 = 0 Now we use the rational root theorem, the possible roots are plus minus 1, 3, 9, 18 we plug in and we get positive 3 as the answer... P.S. Sorry, I don't know latex yet, so I put my solution in a picture

Let $\sqrt [ 3 ]{ 9+4\sqrt { 5 } } +\quad \sqrt [ 3 ]{ 9-4\sqrt [ 2 ]{ 5 } } -x\quad =0\\ \\ consider\\ \quad \\ a=\sqrt [ 3 ]{ 9+4\sqrt { 5 } } ,b=\sqrt [ 3 ]{ 9-4\sqrt { 5 } } ,c=(-x)\\ \\ If\quad a+b+c=0\quad then\quad { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }=3abc\\ \\ { \sqrt [ 3 ]{ 9+4\sqrt { 5 } } }^{ 3 }+{ \sqrt [ 3 ]{ 9-4\sqrt { 5 } } }^{ 3 }+{ (-x) }^{ 3 }=3(\sqrt [ 3 ]{ 9+4\sqrt { 5\quad .9-4\sqrt { 5\quad ) } } } .(-x)\\ \\ 9+4\sqrt { 5 } +9-4\sqrt { 5 } { -x }^{ 3 }\quad =\quad 3\sqrt [ 3 ]{ 81-80 } (-x)\\ \\ 18{ -x }^{ 3 }=\quad 3.1.(-x)\\ \\ 18-{ x }^{ 3 }=-3x\\ \\ { x }^{ 3 }-3x-18=0\\ \\ By\quad factorizing\quad we\quad get,\\ \\ (x-3)({ x }^{ 2 }+3x+6)=0\quad \\ \\ x-3=0\quad \quad { x }^{ 2 }+3x+6=0\quad ,\Delta <0\quad \\ \\ x=3\quad \quad \quad \quad The\quad roots\quad are\quad complex.\\ \\ But\quad the\quad sum\quad of\quad two\quad irrationals\quad cannot\quad be\quad complex\\ \\ \therefore \quad \sqrt [ 3 ]{ 9+4\sqrt { 5 } } +\sqrt [ 3 ]{ 9-4\sqrt { 5 } } =x=3$