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Question Number 151316 by iloveisrael last updated on 20/Aug/21

Find positive integers a and b such that ((a)^(1/3) +(b)^(1/3) −1)^2 = 49+ 20(6)^(1/3)

$$\:\:\:\:\:\:{Find}\:{positive}\:{integers}\: \\ $$$$\:\:\:\:\:{a}\:{and}\:{b}\:{such}\:{that}\: \\ $$$$\:\:\:\:\:\left(\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:−\mathrm{1}\right)^{\mathrm{2}} =\:\mathrm{49}+\:\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$

Answered by Rasheed.Sindhi last updated on 20/Aug/21

((a)^(1/3) +(b)^(1/3) −1)^2 ^(SOLVE for a,b∈Z^+ ) = 49+ 20(6)^(1/3) ((a)^(1/3) +(b)^(1/3) _(m) −1)^2 −49=20(6)^(1/3) ⇒m is not rational (m −1−7)(m −1+7)=20(6)^(1/3) (m−8)(m +6)=1.2^2 .5.(2)^(1/3) .(3)^(1/3) (Obviously, m −8<m+6) perhaps no solution

$$\overset{\mathcal{SOLVE}\:\:{for}\:{a},{b}\in\mathbb{Z}^{+} } {\left(\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:−\mathrm{1}\right)^{\mathrm{2}} }=\:\mathrm{49}+\:\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$$$\left(\underset{{m}} {\underbrace{\sqrt[{\mathrm{3}}]{{a}}\:+\sqrt[{\mathrm{3}}]{{b}}\:}}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{49}=\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}}\: \\ $$$$\Rightarrow{m}\:{is}\:{not}\:{rational} \\ $$$$\left({m}\:−\mathrm{1}−\mathrm{7}\right)\left({m}\:−\mathrm{1}+\mathrm{7}\right)=\mathrm{20}\sqrt[{\mathrm{3}}]{\mathrm{6}} \\ $$$$\left({m}−\mathrm{8}\right)\left({m}\:+\mathrm{6}\right)=\mathrm{1}.\mathrm{2}^{\mathrm{2}} .\mathrm{5}.\sqrt[{\mathrm{3}}]{\mathrm{2}}.\sqrt[{\mathrm{3}}]{\mathrm{3}} \\ $$$$\left({Obviously},\:{m}\:−\mathrm{8}<{m}+\mathrm{6}\right) \\ $$$${perhaps}\:{no}\:{solution} \\ $$

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