One of the real roots of the equation x^3-6x^ | vedclass

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(C) Given equation: $x^3-6x^2+6x-2=0$. Checking option $(a)$: Putting $x=-1$,we get $(-1)^3-6(-1)^2+6(-1)-2 = -1-6-6-2 = -15 
eq 0$. Checking option

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$\frac{2^{1/3}}{2^{1/3}-1}$

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(C) Given equation: $x^3-6x^2+6x-2=0$. Checking option $(a)$: Putting $x=-1$,we get $(-1)^3-6(-1)^2+6(-1)-2 = -1-6-6-2 = -15 
eq 0$. Checking option $(b)$: Putting $x=2$,we get $(2)^3-6(2)^2+6(2)-2 = 8-24+12-2 = -6 
eq 0$. Checking option $(c)$: Let $x = \frac{2^{1/3}}{2^{1/3}-1}$. Applying componendo and dividendo: $\frac{x+1}{x-1} = \frac{2^{1/3}+2^{1/3}-1}{2^{1/3}-2^{1/3}+1} = 2 \cdot 2^{1/3}-1$. $\frac{x+1}{x-1} + 1 = 2 \cdot 2^{1/3}$ $\Rightarrow \frac{2x}{x-1} = 2 \cdot 2^{1/3}$ $\Rightarrow \frac{x}{x-1} = 2^{1/3}$. Cubing both sides: $\frac{x^3}{(x-1)^3} = 2 $ $\Rightarrow x^3 = 2(x^3-3x^2+3x-1) $ $\Rightarrow x^3 = 2x^3-6x^2+6x-2 $ $\Rightarrow x^3-6x^2+6x-2 = 0$. Thus,option $(c)$ is the correct root.

One of the real roots of the equation $x^3-6x^2+6x-2=0$ is

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