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Question Number 181279 by mnjuly1970 last updated on 23/Nov/22

Answered by Rasheed.Sindhi last updated on 25/Nov/22

$$\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0$$$$\overline{)\begin{array}{c}\underset{\Rightarrow a^3+b^3+c^3-3abc=0}{a+b+c=0}\end{array}}$$$$x+1+x+2+x+3-3\sqrt[3]{(x+1)}\sqrt[3]{x+2}\sqrt[3]{x+3}=0$$$$3(x+2)-3\sqrt[3]{(x+1)}\sqrt[3]{x+2}\sqrt[3]{x+3}=0$$$$3{\left(\sqrt[3]{x+2}\right)}^3-3\sqrt[3]{(x+1)}\sqrt[3]{x+2}\sqrt[3]{x+3}=0$$$$3\left(\sqrt[3]{x+2}\right)({\left(\sqrt[3]{x+2}\right)}^2-\sqrt[3]{(x+1)}\sqrt[3]{x+3})=0$$$$\sqrt[3]{x+2}=0\Rightarrow x+2=0\Rightarrow x=-2$$

Answered by Rasheed.Sindhi last updated on 23/Nov/22

$$\sqrt[3]{x+1}+\sqrt[3]{x+2}+\sqrt[3]{x+3}=0$$$$\sqrt[3]{x+1}+\sqrt[3]{x+2}=-\sqrt[3]{x+3}$$$${(\sqrt[3]{x+1}+\sqrt[3]{x+2})}^3={(-\sqrt[3]{x+3})}^3$$$$(x+1)+(x+2)+3\sqrt[3]{(x+1)(x+2)}(\sqrt[3]{x+1}+\sqrt[3]{x+2})=-x-3$$$$3\sqrt[3]{(x+1)(x+2)}(-\sqrt[3]{x+3})=-3x-6$$$$\sqrt[3]{(x+1)(x+2)}\sqrt[3]{x+3})=x+2$$$$(x+1)(x+2)(x+3)={(x+2)}^3$$$$x^3+6x^2+11x+6=x^3+6x^2+12x+8$$$$11x+6=12x+8$$$$x=-2$$$$$$

Commented by Frix last updated on 23/Nov/22

$$\mathrm{generally}\mathrm{in}\mathbb{R}:$$$$\sqrt[3]{p}+\sqrt[3]{q}=\sqrt[3]{r}$$$$p+3\sqrt[3]{p^2}\sqrt[3]{q}+3\sqrt[3]{p}\sqrt[3]{q^2}+q=r^3$$$$p+q+3\sqrt[3]{pq}(\sqrt[3]{p}+\sqrt[3]{q})=r^3$$$$p+q+3\sqrt[3]{pq}\sqrt[3]{r}=r^3$$$$3\sqrt[3]{pqr}=r^3-p-q$$$$27pqr={(r^3-p-q)}^3$$$$$$$$\mathrm{from}\mathrm{your}\mathrm{line}$$$$(x+1)(x+2)(x+3)={(x+2)}^3$$$${\mathrm{it}}^{\prime }\mathrm{s}\mathrm{obvious}\mathrm{that}x+2=0\mathrm{is}\mathrm{a}\mathrm{solution}:$$$$(x+1)(x+2)(x+3)-{(x+2)}^3=0$$$$(x+2)((x+1)(x+3)-{(x+2)}^2)=0$$$$\Rightarrow x_1=-2$$

Commented by Rasheed.Sindhi last updated on 23/Nov/22

$$\mathbb{T}\mathrm{hanks}\mathrm{sir}\mathbb{F}\mathrm{r}\stackrel{⧫}{\mathrm{\Pi }}\mathrm{x}!$$

Answered by Frix last updated on 23/Nov/22

$$‘‘\mathrm{solve}\mathrm{in}{\mathbb{R}}^{″}\iff ‘‘\mathrm{use}\sqrt[3]{-r}=-{\sqrt[3]{r}}^{″}$$$$\Rightarrow x=-2$$$$\mathrm{easy}\mathrm{to}\mathrm{see}:$$$$\sqrt[3]{t-1}+\sqrt[3]{t}+\sqrt[3]{t+1}=0\Rightarrow t=0$$$$t=x+2$$$$x=-2$$

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