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Question Number 141521 by 7770 last updated on 19/May/21

$$\{\begin{array}{l}\sqrt[3]{\mathit{x}+\mathit{y}}+\sqrt{\mathit{x}-3\mathit{y}}=4\\ 2\mathit{x}+3\mathit{y}=17\end{array}$$$$\mathbf{Find}\mathbf{x}\mathbf{and}\mathbf{y}.$$

Answered by MJS_new last updated on 20/May/21

$${\mathrm{let}}^{\prime }\mathrm{s}\mathrm{try}x+y=a^3\wedge x-3y=b^2$$$$\iff$$$$x=\frac{3a^3+b^2}{4}\wedge y=\frac{a^3-b^2}{4}$$$$a+b=4\Rightarrow b=4-a$$$$9a^3-b^2=68\Rightarrow a^3-\frac{1}{9}{a}^2+\frac{8}{9}a-\frac{28}{3}=0\Rightarrow a=2$$$$\Rightarrow b=2$$$$\Rightarrow x=7\wedge y=1$$

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