Question Number 160806 by HongKing last updated on 06/Dec/21
Answered by MJS_new last updated on 06/Dec/21
$${A}=\mathrm{0}\vee{A}=\mathrm{5} \\ $$$$ \\ $$$${y}={A}−{x} \\ $$$${x}^{\mathrm{3}} +\left({A}−{x}\right)^{\mathrm{3}} =\mathrm{5}{x}\left({A}−{x}\right) \\ $$$$\left(\mathrm{3}{A}+\mathrm{5}\right){x}^{\mathrm{2}} −{A}\left(\mathrm{3}{A}+\mathrm{5}\right){x}+{A}^{\mathrm{3}} =\mathrm{0} \\ $$$${x}=\frac{{A}}{\mathrm{2}}\pm\frac{{A}\sqrt{\mathrm{5}−{A}}}{\mathrm{2}\sqrt{\mathrm{3}{A}+\mathrm{5}}} \\ $$$$\mathrm{one}\:\mathrm{solution}\:\mathrm{if}\:\frac{{A}\sqrt{\mathrm{5}−{A}}}{\mathrm{2}\sqrt{\mathrm{3}{A}+\mathrm{5}}}=\mathrm{0}\:\Rightarrow\:{A}=\mathrm{0}\vee{A}=\mathrm{5} \\ $$
Commented by HongKing last updated on 06/Dec/21
$$\mathrm{cool}\:\mathrm{thank}\:\mathrm{you}\:\mathrm{so}\:\mathrm{much}\:\mathrm{my}\:\mathrm{dear}\:\mathrm{Sir} \\ $$