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Question Number 200657 by cherokeesay last updated on 21/Nov/23

Answered by Frix last updated on 21/Nov/23

3((27x^2 +24x+((28)/3)))^(1/4) =4+((((81x)/2)−1))^(1/3)   3(((81x^2 +72x+28)/3))^(1/4) =4+(((81x−2)/2))^(1/3)   x=(t/9)  3(((t^2 +8t+28)/3))^(1/4) =4+(((9t−2)/2))^(1/3)   Now I′m looking for both roots to vanish.  Starting with ((9t−2)/2)=n^3  ⇔ 9t=2(n^3 +1).  Very obviously t=n=2 is a solution.  3(((2^2 +8×2+28)/3))^(1/4) =4+(((9×2−2)/2))^(1/3)   3(((48)/3))^(1/4) =4+(((16)/2))^(1/3)   3×((16))^(1/4) =4+(8)^(1/3)   3×2=4+2  True  ⇒ x=(2/9)

$$\mathrm{3}\sqrt[{\mathrm{4}}]{\mathrm{27}{x}^{\mathrm{2}} +\mathrm{24}{x}+\frac{\mathrm{28}}{\mathrm{3}}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{81}{x}}{\mathrm{2}}−\mathrm{1}} \\ $$$$\mathrm{3}\sqrt[{\mathrm{4}}]{\frac{\mathrm{81}{x}^{\mathrm{2}} +\mathrm{72}{x}+\mathrm{28}}{\mathrm{3}}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{81}{x}−\mathrm{2}}{\mathrm{2}}} \\ $$$${x}=\frac{{t}}{\mathrm{9}} \\ $$$$\mathrm{3}\sqrt[{\mathrm{4}}]{\frac{{t}^{\mathrm{2}} +\mathrm{8}{t}+\mathrm{28}}{\mathrm{3}}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}{t}−\mathrm{2}}{\mathrm{2}}} \\ $$$$\mathrm{Now}\:\mathrm{I}'\mathrm{m}\:\mathrm{looking}\:\mathrm{for}\:\mathrm{both}\:\mathrm{roots}\:\mathrm{to}\:\mathrm{vanish}. \\ $$$$\mathrm{Starting}\:\mathrm{with}\:\frac{\mathrm{9}{t}−\mathrm{2}}{\mathrm{2}}={n}^{\mathrm{3}} \:\Leftrightarrow\:\mathrm{9}{t}=\mathrm{2}\left({n}^{\mathrm{3}} +\mathrm{1}\right). \\ $$$$\mathrm{Very}\:\mathrm{obviously}\:{t}={n}=\mathrm{2}\:\mathrm{is}\:\mathrm{a}\:\mathrm{solution}. \\ $$$$\mathrm{3}\sqrt[{\mathrm{4}}]{\frac{\mathrm{2}^{\mathrm{2}} +\mathrm{8}×\mathrm{2}+\mathrm{28}}{\mathrm{3}}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{9}×\mathrm{2}−\mathrm{2}}{\mathrm{2}}} \\ $$$$\mathrm{3}\sqrt[{\mathrm{4}}]{\frac{\mathrm{48}}{\mathrm{3}}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\frac{\mathrm{16}}{\mathrm{2}}} \\ $$$$\mathrm{3}×\sqrt[{\mathrm{4}}]{\mathrm{16}}=\mathrm{4}+\sqrt[{\mathrm{3}}]{\mathrm{8}} \\ $$$$\mathrm{3}×\mathrm{2}=\mathrm{4}+\mathrm{2} \\ $$$$\mathrm{True} \\ $$$$\Rightarrow\:{x}=\frac{\mathrm{2}}{\mathrm{9}} \\ $$

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