Cube to the root!

Let $\color{#D61F06}{A=\sqrt[3]{45+29\sqrt2}}$ , $\color{#3D99F6}{B=\sqrt[3]{45-29\sqrt2}}$ and $\color{#20A900}S=\color{#D61F06}A+\color{#3D99F6}B=\color{#D61F06}{\sqrt[3]{45+29\sqrt2}}+\color{#3D99F6}{\sqrt[3]{45-29\sqrt2}}$

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,

$\begin{aligned}\color{#302B94}{A^3+B^3}=&\ \left(\color{#D61F06}{\sqrt[3]{45+29\sqrt2}}\right)^3+\left(\color{#3D99F6}{\sqrt[3]{45-29\sqrt2}}\right)^3\\=&\ 45+29\sqrt2+45-29\sqrt2\\=&\ \color{#302B94}{90}\end{aligned}$

$\begin{aligned}\color{magenta}{3AB}=&\ 3\left(\color{#D61F06}{\sqrt[3]{45+29\sqrt2}}\right)\left(\color{#3D99F6}{\sqrt[3]{45-29\sqrt2}}\right)\\=&\ 3\sqrt[3]{(45+29\sqrt2)(45-29\sqrt2)}\\=&\ 3\sqrt[3]{2025-1682}\\=&\ 3\sqrt[3]{343}\\=&\ 3(7)\\=&\ \color{magenta}{21}\end{aligned}$

Instead $\color{#302B94}{A^3+B^3}$ with $\color{#302B94}{90}$ and $\color{magenta}{3AB}$ with $\color{magenta}{21}$ in equation $(*)$

$\begin{aligned}\color{#20A900}S^3=&\ \color{#302B94}{90}+\color{magenta}{21}\color{#20A900}S\\\color{#20A900}S^3-21\color{#20A900}S-90=&\ 0\\(\color{#20A900}S-6)(\color{#20A900}S^2+6\color{#20A900}S+15)=&\ 0\\\color{#20A900}{S=}&\ \color{#20A900}6\end{aligned}$

From equation

$\begin{aligned}\color{#20A900}S^2+6\color{#20A900}S+15=&\ 0\\\Delta=&\ 6^2-4(1)(15)=-24\end{aligned}$

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

Hence, $\sqrt[3]{45+29\sqrt2}+\sqrt[3]{45-29\sqrt2}=\boxed6$

Assume that $45 \pm 29\sqrt{2}$ takes the form of $(a \pm \sqrt{2})^3$ , then we have:

$\begin{aligned} (a \pm \sqrt{2})^3 & = \color{#D61F06}{a^3} \pm \color{#3D99F6}{(3a^2)}\sqrt{2} + \color{#D61F06}{6a} \pm \color{#3D99F6}{2}\sqrt{2} \\ & = \color{#D61F06}{45} \pm \color{#3D99F6}{29} \sqrt{2} \quad \quad \quad \color{#3D99F6}{\text{Equating multiples of }\sqrt{2} \text{ on both sides}} \\ \Rightarrow \color{#3D99F6}{3a^2 + 2} & = \color{#3D99F6}{29} \\ \Rightarrow a & = 3 \quad \quad \quad \quad \quad \quad \quad \color{#3D99F6}{\text{Checking the other number}} \\ \color{#D61F06}{a^3 + 6a} & = 27+18 = \color{#D61F06}{45} \quad \quad \color{#3D99F6}{\text{Correct!}} \end{aligned}$

Therefore,

$\begin{aligned} \sqrt[3]{49+29\sqrt{2}} + \sqrt[3]{49-29\sqrt{2}} & = \sqrt[3]{(3+\sqrt{2})^3} + \sqrt[3]{(3-\sqrt{2})^3} \\ & = 3+\sqrt{2} + 3-\sqrt{2} \\ &= \boxed{6} \end{aligned}$

Let x=a+b Raise both sides to 3 we get (x)³ = (a+b)³ ---> x³=a³+b³+3ab(a+b) ----> x³=90+3(7)(x) Solve for x we get x=6