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Question Number 18748 by khamizan833@yahoo.com last updated on 29/Jul/17

Answered by Joel577 last updated on 29/Jul/17

f(x) = (x − 1)^3 − 6x f(1 + (2)^(1/3) + (4)^(1/3) ) = ((2)^(1/3) + (4)^(1/3) )^3 − 6(1 + (2)^(1/3) + (4)^(1/3) ) = (2 + 3((16))^(1/3) + 3((32))^(1/3) + 4) − (6 + 6(2)^(1/3) + 6(4)^(1/3) ) = 6 + 6(2)^(1/3) + 6(4)^(1/3) − 6 − 6(2)^(1/3) − 6(4)^(1/3) = 0

$${f}\left({x}\right)\:=\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}} \:−\:\mathrm{6}{x} \\ $$$${f}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)\:=\:\left(\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right)^{\mathrm{3}} \:−\:\mathrm{6}\left(\mathrm{1}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\mathrm{2}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{16}}\:+\:\mathrm{3}\sqrt[{\mathrm{3}}]{\mathrm{32}}\:+\:\mathrm{4}\right)\:−\:\left(\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{6}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:+\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}}\:−\:\mathrm{6}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{2}}\:−\:\mathrm{6}\sqrt[{\mathrm{3}}]{\mathrm{4}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\mathrm{0} \\ $$

Commented by khamizan833@yahoo.com last updated on 30/Jul/17

Thank you sir

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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