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Question Number 147673 by mnjuly1970 last updated on 22/Jul/21

Answered by Rasheed.Sindhi last updated on 22/Jul/21

$$\sqrt[3]{x}+\sqrt[3]{x-2}=\sqrt[3]{x-1}$$$${(\sqrt[3]{x}+\sqrt[3]{x-2})}^3={\left(\sqrt[3]{x-1}\right)}^3$$$$x+x-2+3\sqrt[3]{x}\sqrt[3]{x-2}(\sqrt[3]{x}+\sqrt[3]{x-2})=x-1$$$$3\sqrt[3]{x}\sqrt[3]{x-2}\left(\sqrt[3]{x-1}\right)=-x+1$$$$27\left(x\right)(x-1)(x-2)=-{(x-1)}^3$$$$27\left(x\right)(x-1)(x-2)+{(x-1)}^3=0$$$$(x-1)\{27\left(x\right)(x-2)+{(x-1)}^2\}=0$$$$x=1\mid 27\left(x\right)(x-2)+{(x-1)}^2=0$$$$27x^2-54x+x^2-2x+1=0$$$$28x^2-56x+1=0$$$$x=\frac{56\pm \sqrt{{56}^2-4\left(28\right)}}{56}$$$$=\frac{56\pm 12\sqrt{21}}{56}=\frac{14\pm 3\sqrt{21}}{14}$$$$a=1,b=\frac{14+3\sqrt{21}}{14},c=\frac{14-3\sqrt{21}}{14}$$$$\sqrt[3]{\frac{a}{a-2}}=\sqrt[3]{\frac{1}{1-2}}=-1$$$$\sqrt[3]{\frac{b}{b-2}}=\sqrt[3]{\frac{\frac{14+3\sqrt{21}}{14}}{\frac{14+3\sqrt{21}}{14}-2}}=\sqrt[3]{\frac{14+3\sqrt{21}}{14+3\sqrt{21}-28}}$$$$=\sqrt[3]{\frac{14+3\sqrt{21}}{-14+3\sqrt{21}}}=\frac{{(14+3\sqrt{21})}^{2/3}}{{\{196-9\left(21\right)\}}^{1/3}}$$$$=\frac{{(14+3\sqrt{21})}^{2/3}}{{\{196-9\left(21\right)\}}^{1/3}}=\frac{{(14+3\sqrt{21})}^{2/3}}{7^{1/3}}$$$$\sqrt[3]{\frac{c}{c-2}}=\sqrt[3]{\frac{14-3\sqrt{21}}{-14-3\sqrt{21}}}$$$$=\frac{{(14-3\sqrt{21})}^{2/3}}{{\{196-9\left(21\right)\}}^{1/3}}=\frac{{(14-3\sqrt{21})}^{2/3}}{7^{1/3}}$$$$\sqrt[3]{\frac{a}{a-2}}+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}$$$$=-1+\frac{{(14+3\sqrt{21})}^{2/3}}{7^{1/3}}+\frac{{(14-3\sqrt{21})}^{2/3}}{7^{1/3}}$$$$=-1+\frac{{(385+84\sqrt{21})}^{1/3}}{7^{1/3}}+\frac{{(385-84\sqrt{21})}^{1/3}}{7^{1/3}}$$

Commented by mr W last updated on 22/Jul/21

$$=\sqrt[3]{\frac{14+3\sqrt{21}}{-14+3\sqrt{21}}}=\frac{{(14+3\sqrt{21})}^{2/3}}{{\{-196+9\left(21\right)\}}^{1/3}}$$$$....$$$$=-1-\frac{{(385+84\sqrt{21})}^{1/3}}{7^{1/3}}-\frac{{(385-84\sqrt{21})}^{1/3}}{7^{1/3}}$$$$=-6$$

Commented by Rasheed.Sindhi last updated on 22/Jul/21

$$\mathbf{Sir}{\mathrm{how}}^{\prime }\mathrm{s}\mathrm{the}\mathrm{middle}\mathrm{line}\mathrm{equal}\mathrm{to}-6?$$

Commented by mr W last updated on 22/Jul/21

$$=-1-\frac{{(385+84\sqrt{21})}^{1/3}}{7^{1/3}}-\frac{{(385-84\sqrt{21})}^{1/3}}{7^{1/3}}$$$$=-1-{(55+12\sqrt{21})}^{1/3}-{(55-12\sqrt{21})}^{1/3}$$$$a={(55+12\sqrt{21})}^{1/3}$$$$b={(55-12\sqrt{21})}^{1/3}$$$$a^3+b^3=110$$$$ab=1$$$${(a+b)}^3=a^3+b^3+3ab(a+b)$$$${(a+b)}^3=110+3(a+b)$$$${(a+b)}^3-3(a+b)-110=0$$$$u=a+b$$$$(u-5)(u^2+5u+22)=0$$$$\Rightarrow a+b=u=5$$$$\Rightarrow {(55+12\sqrt{21})}^{1/3}+{(55-12\sqrt{21})}^{1/3}=5$$

Commented by Rasheed.Sindhi last updated on 23/Jul/21

$$\mathrm{A}\mathrm{bundle}\mathrm{of}\mathrm{thanks}\mathrm{sir}!$$

Commented by mnjuly1970 last updated on 23/Jul/21

$$grateful...$$

Answered by mr W last updated on 22/Jul/21

$$\sqrt[3]{\frac{x}{x-2}}+1=\sqrt[3]{\frac{x-1}{x-2}}$$$$\sqrt[3]{\frac{x}{x-2}}+1=\sqrt[3]{\frac{x-1}{x}}\sqrt[3]{\frac{x}{x-2}}$$$$(\sqrt[3]{\frac{x-1}{x}}-1)\sqrt[3]{\frac{x}{x-2}}=1$$$$lett=\sqrt[3]{\frac{x}{x-2}}$$$$\Rightarrow x=\frac{2t^3}{t^3-1}$$$$\Rightarrow x-1=\frac{t^3+1}{t^3-1}$$$$(\sqrt[3]{\frac{t^3+1}{2t^3}}-1)t=1$$$$\sqrt[3]{\frac{t^3+1}{2}}=t+1$$$$\Rightarrow t^3+1=2(t^3+3t^2+3t+1)$$$$\Rightarrow t^3+6t^2+6t+1=0$$$$\Rightarrow \underset{i=1}{\sum ^3}{t}_i=-6$$$$i.e.\underset{i=1}{\sum ^3}\sqrt[3]{\frac{x_i}{x_i-2}}=-6$$$$i.e.\sqrt[3]{\frac{a}{a-2}}+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}=-6$$

Commented by Rasheed.Sindhi last updated on 22/Jul/21

$$\vee \cap \stackrel{\bullet }{\mid }\subset \in \mathrm{\$}\mathrm{ir}!$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Sir}\mathrm{mrW}\mathrm{please}\mathrm{tag}\mathrm{the}\mathrm{general}\mathrm{symmetry}\mathrm{polynomial}\mathrm{you}\mathrm{did}$$$$\mathrm{sometimes}\mathrm{ago}.$$$$\mathrm{like},$$$$\mathrm{x}+\mathrm{y}+\mathrm{z}=1$$$${\mathrm{x}}^2+{\mathrm{y}}^2+{\mathrm{z}}^2=2$$$${\mathrm{x}}^3+{\mathrm{y}}^3+{\mathrm{z}}^3=3$$$$\mathrm{then},{\mathrm{x}}^5+{\mathrm{y}}^5+{\mathrm{z}}^5=??$$$$$$$$\mathrm{Just}\mathrm{tag}\mathrm{the}\mathrm{one}\mathrm{you}\mathrm{did}\mathrm{sometimes}\mathrm{ago}.$$$$\mathrm{Thanks}\mathrm{sir}.\mathrm{God}\mathrm{bless}\mathrm{you}.$$

Commented by Tawa11 last updated on 23/Jul/21

$$\mathrm{Seen}\mathrm{sir}.\mathrm{Thanks}.$$

Commented by mnjuly1970 last updated on 23/Jul/21

$$thanksalotmrW$$

Answered by Rasheed.Sindhi last updated on 23/Jul/21

$$\sqrt[3]{x}+\sqrt[3]{x-2}=\sqrt[3]{x-1}$$$$\sqrt[3]{\frac{a}{a-2}}+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}=?$$$$Letx-1=y^3\Rightarrow x=y^3+1$$$$\sqrt[3]{y^3+1}+\sqrt[3]{y^3-1}=y$$$$Cubingbothsides:$$$$y^3+1+y^3-1+3\sqrt[3]{(y^3+1)(y^3-1}\left(y\right)=y^3$$$$2y^3+3y\sqrt[3]{y^6-1}=y^3$$$$y^3+3y\sqrt[3]{y^6-1}=0$$$$y(y^2+3\sqrt[3]{y^6-1})=0$$$$y=0\mid y^2+3\sqrt[3]{y^6-1}=0$$$$x-1=0\mid y^2=-3\sqrt[3]{y^6-1}$$$$x=1\mid y^6=-27(y^6-1)$$$$28y^6=27\Rightarrow {\left(y^3\right)}^2=\frac{27}{28}$$$${(x-1)}^2=\frac{27}{28}$$$$28x^2-56x+1=0$$$$x=\frac{56\pm \sqrt{{56}^2-4\left(28\right)}}{56}$$$$x=\frac{56\pm 12\sqrt{21}}{56}=\frac{14\pm 3\sqrt{21}}{14}$$$$a=1,b=\frac{14+3\sqrt{21}}{14},c=\frac{14-3\sqrt{21}}{14}$$$$A=\sqrt[3]{\frac{a}{a-2}}+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}$$$$A=\sqrt[3]{\frac{1}{1-2}}+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}$$$$A=-1+\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}}$$$${(A+1)}^3={(\sqrt[3]{\frac{b}{b-2}}+\sqrt[3]{\frac{c}{c-2}})}^3$$$$=\frac{b}{b-2}+\frac{c}{c-2}+3\sqrt[3]{\frac{bc}{bc-2(b+c)+4}}(A+1)$$$$=\frac{bc-2b+bc-2c}{bc-2(b+c)+4}+3\sqrt[3]{\frac{bc}{bc-2(b+c)+4}}(A+1)$$$$=\frac{2bc-2(b+c)}{bc-2(b+c)+4}+3\sqrt[3]{\frac{bc}{bc-2(b+c)+4}}(A+1)$$$$bc=\left(\frac{14+3\sqrt{21}}{14}\right)\left(\frac{14-3\sqrt{21}}{14}\right)=\frac{196-9\left(21\right)}{196}$$$$=1/28$$$$b+c=\frac{14+3\sqrt{21}}{14}+\frac{14-3\sqrt{21}}{14}=2$$$${(A+1)}^3=\frac{2\left(1/28\right)-2\left(2\right)}{\left(1/28\right)-2\left(2\right)+4}+3\sqrt[3]{\frac{1/28}{\left(1/28\right)-2\left(2\right)+4}}(A+1)$$$${(A+1)}^3=\frac{\frac{1}{14}-4}{\frac{1}{28}}+3\sqrt[3]{\frac{1/28}{\left(1/28\right)}}(A+1)$$$${(A+1)}^3=\frac{\frac{-55}{14}}{\frac{1}{28}}+3\sqrt[3]{\frac{1/28}{\left(1/28\right)}}(A+1)$$$$=-110+3A+3$$$${(A+1)}^3-3A+107=0$$$$A^3+1+3A(A+1)-3A+107=0o$$$$A^3+3A^2+3A-3A+108=0$$$$A^3+3A^2+108=0$$$$(A+6)(A^2-3A+18)=0$$$$A=-6,\frac{3\pm \sqrt{9-72}}{2}=\frac{3\pm 3i\sqrt{7}}{2}(Rejected$$$$becauseAisobviouslyreal)$$

Commented by mr W last updated on 23/Jul/21

$$goodtoo!$$

Commented by Rasheed.Sindhi last updated on 23/Jul/21

$$\mathcal{A}nalternate$$$$although$$$$somewhatmoretimeconsuming$$$$yet$$$$notbadforpractising.$$

Commented by Rasheed.Sindhi last updated on 23/Jul/21

$$\mathcal{T}hank\mathcal{Y}ou$$$$\mathit{s}\mathit{i}\mathit{r}mrW!$$

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