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Question Number 43793 by gunawan last updated on 15/Sep/18

$$\mathrm{The}\mathrm{value}\mathrm{of}{}^3\sqrt{20+14\sqrt{2}}{+}^3\sqrt{20-14\sqrt{2}}\mathrm{is}\dots$$

Answered by Joel578 last updated on 15/Sep/18

$$\mathrm{Let}a=20+14\sqrt{2}$$$$b=20-14\sqrt{2}$$$$ab=(20+14\sqrt{2})(20-14\sqrt{2})=400-392=8$$$$$$$$\sqrt[3]{a}+\sqrt[3]{b}=x$$$$a+3\sqrt[3]{a^{2}b}+3\sqrt[3]{{ab}^2}+b=x^3$$$$(a+b)+3\sqrt[3]{ab}(\sqrt[3]{a}+\sqrt[3]{b})=x^3$$$$\left(40\right)+3\sqrt[3]{8}\left(x\right)=x^3$$$$40+6x=x^3$$$$x^3-6x-40=0$$$$(x-4)(x^2+4x+10)=0$$$$\therefore x=4$$

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Sep/18

$$20+14\sqrt{2}$$$$=2^3+3.2^2.\left(\sqrt{2}\right)+3.2.{\left(\sqrt{2}\right)}^2+{\left(\sqrt{2}\right)}^3$$$$={(2+\sqrt{2})}^3$$$$20-14\sqrt{2}={(2-\sqrt{2})}^3$$$${(20+14\sqrt{2})}^{\frac{1}{3}}+{(20-14\sqrt{2})}^{\frac{1}{3}}$$$$=2+\sqrt{2}+2-\sqrt{2}$$$$=4$$

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