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Question Number 51772 by Cheyboy last updated on 30/Dec/18

$${\mathrm{x}}^3+12\mathrm{x}+12=0$$$$$$$$\mathrm{Express}\mathrm{your}\mathrm{answer}\mathrm{in}\mathrm{surd}\mathrm{form}$$$$$$

Answered by ajfour last updated on 30/Dec/18

$$x={(-6+\sqrt{36+64})}^{1/3}-{(6+\sqrt{36+64})}^{1/3}$$$$=\sqrt[3]{4}-\sqrt[3]{16}.$$

Commented by Cheyboy last updated on 30/Dec/18

$$\mathrm{Thank}\mathrm{you}\mathrm{sir}.\mathrm{but}\mathrm{i}\mathrm{would}\mathrm{appreciate}$$$$\mathrm{to}\mathrm{know}\mathrm{the}\mathrm{working}$$

Answered by mr W last updated on 31/Dec/18

$$x^3+12x+12=0$$$$cubiceqn.ofthiskindcanbesolved$$$$infollowingway:$$$$$$$$weknowthat$$$$u^3+v^3=(u+v)(u^2-uv+v^2)=(u+v)[{(u+v)}^2-3uv]$$$$\Rightarrow {(u+v)}^3-3uv(u+v)-(u^3+v^3)=0$$$$$$$$whenyoucomparethiswithoureqn.$$$$youcanseethat$$$$ifweput3uv=-12,u^3+v^3=-12,$$$$thenu+visthesolutionofoureqn.$$$$$$$$3uv=-12\Rightarrow v=-\frac{4}{u}$$$$u^3+v^3=-12\Rightarrow u^3-\frac{4^3}{u^3}+12=0$$$$\Rightarrow u^6+12u^3-64=0(quadraticeqn.for{u}^3)$$$$\Rightarrow (u^3+16)(u^3-4)=0$$$$\Rightarrow u^3=4(-16willalsodo,butgivesthesame)$$$$\Rightarrow u=\sqrt[3]{4}$$$$\Rightarrow v=-\frac{4}{\sqrt[3]{4}}=-2\sqrt[3]{2}$$$$\Rightarrow u+v=\sqrt[3]{4}-2\sqrt[3]{2}$$$$$$$$i.e.thesolutionof{x}^3+12x+12=0is$$$$x=\sqrt[3]{4}-2\sqrt[3]{2}\approx -0.9324$$

Commented by Cheyboy last updated on 30/Dec/18

$$\mathrm{Thank}\mathrm{you}\mathrm{very}\mathrm{much}\mathrm{sir}$$$$$$

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