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Question Number 154677 by mathdanisur last updated on 20/Sep/21

$$\mathrm{Solve}\mathrm{for}\mathrm{real}\mathrm{numbers}:$$$$\frac{{\mathrm{x}}^9-{256\mathrm{x}}^3-791}{{84\mathrm{x}}^3}=\sqrt[3]{4\mathrm{x}+7}$$

Commented by MJS_new last updated on 20/Sep/21

$$\frac{x^9-256x-791}{84x^3}=\sqrt[3]{4x+7}$$$$\mathrm{we}\mathrm{can}\mathrm{solve}\mathrm{it}...$$$$\mathrm{let}x=\sqrt[3]{4x+7}\Rightarrow x^3-4x-7=0\left(1\right)$$$$\Rightarrow$$$$\frac{x^9-256x-791}{84x^3}=x$$$$\iff$$$$x^9-84x^4-256x-791=0\left(2\right)$$$$(x^3-4x-7)(x^6+4x^4+7x^3+16x^2-28x+113)=0$$$$\Rightarrow$$$$\mathrm{the}\mathrm{solutions}\mathrm{of}\left(1\right)\mathrm{also}\mathrm{solve}\left(2\right)$$$$x\in \mathbb{R}\Rightarrow x=\sqrt[3]{\frac{7}{2}-\frac{\sqrt{3201}}{18}}+\sqrt[3]{\frac{7}{2}+\frac{\sqrt{3201}}{18}}$$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{There}\mathrm{was}\mathrm{a}\mathrm{typo}.\mathrm{Thank}\mathrm{you}\mathrm{for}\mathrm{your}\mathrm{attention}\mathbf{S}\mathrm{er}.$$$$\mathrm{The}\mathrm{correct}\mathrm{is}...256\mathrm{x},\mathrm{not}{256\mathrm{x}}^3.$$$$\mathrm{Sorry}\mathrm{for}\mathrm{my}\mathrm{mistake}\mathbf{S}\mathrm{er}.$$

Commented by MJS_new last updated on 20/Sep/21

$$\mathrm{only}\mathrm{approximation}\mathrm{is}\mathrm{possible}$$$$\mathrm{I}\mathrm{get}$$$$x_1\approx -2.05382774$$$$x_2\approx -1.30927779$$$$x_3\approx 2.82876252$$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{creativ}\mathrm{solution},\mathrm{thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

Commented by mr W last updated on 20/Sep/21

$${it}^{\prime }seachtimeawondertomehow$$$$youget$$$$(x^3-4x-7)(x^6+4x^4+7x^3+16x^2-28x+113)=0$$$$from$$$$x^9-84x^4-256x-791=0.$$$$great!$$

Commented by mathdanisur last updated on 20/Sep/21

$$\mathrm{Thank}\mathrm{you}\mathrm{so}\mathrm{much}\mathbf{S}\mathrm{er}$$

Commented by MJS_new last updated on 20/Sep/21

$$\mathrm{this}\mathrm{time}\mathrm{I}\mathrm{first}\mathrm{plotted}\mathrm{both}\mathrm{sides}\mathrm{of}\mathrm{the}$$$$\mathrm{given}\mathrm{equation}\mathrm{and}\mathrm{it}\mathrm{looked}\mathrm{as}\mathrm{if}\mathrm{the}\mathrm{only}$$$$\mathrm{solution}\mathrm{was}\mathrm{on}\mathrm{the}\mathrm{line}y=x.\mathrm{then}\mathrm{I}\mathrm{tried}\mathrm{to}$$$$\mathrm{verify}\mathrm{this}\mathrm{and}\mathrm{it}\mathrm{was}\mathrm{easy}\mathrm{to}\mathrm{divide}$$$$(x^9-84x^4-256x-791)/(x^3-4x-7)$$$$\mathrm{I}\mathrm{see}\mathrm{no}\mathrm{chance}\mathrm{to}\mathrm{get}\mathrm{the}\mathrm{exact}\mathrm{result}$$$$\mathrm{without}\mathrm{this}\mathrm{very}\mathrm{lucky}\mathrm{coincidence}...$$

Commented by MJS_new last updated on 21/Sep/21

$$\mathrm{we}\mathrm{can}\mathrm{build}\mathrm{similar}\mathrm{questions}\mathrm{this}\mathrm{way}:$$$$\frac{x^9-a^{4}x-(a^3+b^2)b}{3{abx}^3}=\sqrt[3]{ax+b}$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{more}\mathrm{a}\mathrm{magic}\mathrm{trick}\mathrm{than}\mathrm{serious}\mathrm{math}$$$$\mathrm{let}\sqrt[3]{ax+b}=x\iff x^3-ax-b=0\left(1\right)$$$$\Rightarrow$$$$\frac{x^9-a^{4}x-(a^3+b^2)b}{3{abx}^3}=x$$$$x^9-4{abx}^4-a^{4}x-(a^3+b^2)b=0\left(2\right)$$$$(x^3-ax-b)(x^6+{ax}^4+{bx}^3+a^2{x}^2-abx+a^3+b^2)=0$$$$\mathrm{obviously}\mathrm{the}\mathrm{real}\mathrm{solution}\left(\mathrm{s}\right)\mathrm{of}\left(1\right)\mathrm{also}$$$$\mathrm{solve}\left(2\right).$$

Commented by MJS_new last updated on 21/Sep/21

$$\mathrm{question}154303\mathrm{works}\mathrm{exactly}\mathrm{the}\mathrm{same}\mathrm{way}$$

Commented by mathdanisur last updated on 21/Sep/21

$$\mathrm{Very}\mathrm{nice}\mathbf{S}\mathrm{er},\mathrm{thank}\mathrm{you}$$

Answered by imjagoll last updated on 21/Sep/21

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