Question

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

• A. $-1$
• B. $2$
• C. $\frac{2^{\frac{1}{3}}}{2^{\frac{1}{3}}-1}$
• D. $\frac{2^{\frac{1}{3}}}{2^{\frac{1}{3}}+1}$

Answer 1

Answer: (C)

We have,
$x^{3}-6 x^{2}+6 x-2=0$
Let $f(x)=x^{3}-6 x^{2}+6 x-2$
$f(4)=64-96+24-2=-10<0$
$f(5)=125-150+30-2=3>0$
$\therefore $ Root lie between 4 and 5 .
Hence, $\frac{2^{1 / 3}}{2^{1 / 3}-1}$ is lie between 4 and 5 .
$\therefore \frac{2^{\frac{1}{3}}}{2^{\frac{1}{3}}-1}$ is one real root of $x^{3}-6 x^{2}+6 x-2=0$