Solución problema 70

Elevando al cubo

$$N^3=45+29\sqrt{2}+3\sqrt[3]{(45+29\sqrt{2}{)}^2(45-29\sqrt{2})}+\sqrt[3]{\left(45+29\sqrt{2}\right)(45-29\sqrt{2}{)}^2}+45-29\sqrt{2}=$$

$$90+3\sqrt[3]{\left(45+29\sqrt{2}\right)\left(45+29\sqrt{2}\right)\left(45-29\sqrt{2}\right)}+3\sqrt[3]{\left(45+29\sqrt{2}\right)\left(45-29\sqrt{2}\right)\left(45-29\sqrt{2}\right)}=$$

$$90+3\sqrt[3]{\left(45+29\sqrt{2}\right)343}+3\sqrt[3]{\left(45-29\sqrt{2}\right)343}=90+21\sqrt[3]{45+29\sqrt{2}}+21\sqrt[3]{45-29\sqrt{2}}=$$

$$90+21\left(\sqrt[3]{45+29\sqrt{2}}+\sqrt[3]{45-29\sqrt{2}}\right)=90+21N$$

Por tanto $N$ es la solución de la ecuación $N^3-21N-90=0$. Esta exuación tiene como soluciones N=6, y otras dos raíces complejas no reales. Por tanto $N=6$