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Question Number 99089 by bramlex last updated on 18/Jun/20
((9+((9+((9+((9+...))^(1/(3  )) ))^(1/(3  )) ))^(1/(3  )) ))^(1/(3  )) −(√(8−(√(8−(√(8−(√(8−...))))))))
$$\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+\sqrt[{\mathrm{3}\:\:}]{\mathrm{9}+…}}}}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−…}}}} \\ $$
Answered by MJS last updated on 18/Jun/20
a=((9+((9+...))^(1/3) ))^(1/3)   a^3 =9+a  a^3 −a−9=0  a=(((9/2)+((√(6549))/(18))))^(1/3) +(((9/2)−((√(6549))/(18))))^(1/3)   b=(√(8−(√(8−...))))  b^2 =8−b  b=−(1/2)+((√(33))/2)  a−b=(((9/2)+((√(6549))/(18))))^(1/3) +(((9/2)−((√(6549))/(18))))^(1/3) +((1−(√(33)))/2)
$${a}=\sqrt[{\mathrm{3}}]{\mathrm{9}+\sqrt[{\mathrm{3}}]{\mathrm{9}+…}} \\ $$$${a}^{\mathrm{3}} =\mathrm{9}+{a} \\ $$$${a}^{\mathrm{3}} −{a}−\mathrm{9}=\mathrm{0} \\ $$$${a}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}} \\ $$$${b}=\sqrt{\mathrm{8}−\sqrt{\mathrm{8}−…}} \\ $$$${b}^{\mathrm{2}} =\mathrm{8}−{b} \\ $$$${b}=−\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$$${a}−{b}=\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}+\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}}{\mathrm{2}}−\frac{\sqrt{\mathrm{6549}}}{\mathrm{18}}}+\frac{\mathrm{1}−\sqrt{\mathrm{33}}}{\mathrm{2}} \\ $$
Commented by bemath last updated on 18/Jun/20
cardano
$$\mathrm{cardano} \\ $$

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