Question

please answer this question

Answer 1

Q : If
$x = 2 + {2}^{ \frac{1}{3} } + {2}^{ \frac{2}{3} }$
then find the value of
${x}^{3} - 6 {x}^{2} + 6x \: \: \: \: is \:$
Final Answer : (b) 2

Solution.:


$x \: = 2 + {2}^{ \frac{1}{3} } + {2}^{ \frac{2}{3} } \\ = > (x - 2) = {2}^{ \frac{1}{3} } (1 + {2}^{ \frac{1}{3} } ) \\ = > {(x - 2)}^{3} = 2 {(1 + {2}^{ \frac{1}{3} } )}^{3} \\ = > {x}^{3} - 8 - 6x(x - 2) = \\ 2(1 + 2 + 3 \times {2}^{ \frac{1}{3} } (1 + {2}^{ \frac{1}{3} } )) \\ = > {x}^{3} - 8 - 6 {x}^{2} + 12x = \\ 6(1 + {2}^{ \frac{1}{3} } + {2}^{ \frac{2}{3} } ) \\ = > {x}^{3} - 8 - 6 {x}^{2} + 12x = 6(x - 1) \\ = > {x}^{3} - 8 - 6 {x}^{2} + 12x - 6x + 6 = 0 \\ = > {x}^{3} - 6 {x}^{2} + 6x = 8 - 6 = 2$