Calculate the value

$\right)$ ) $x^{3}-6x^{2}+6x-2=0$ $x=2+2^{\dfrac {2} {3}}+2^{\dfrac {1} {3}}$ 


Question $4$ 
If $x$ is $6$ what $|s$ 
$\dfrac {1} {3}x$ 
$1frac\left(1\right)\left(3\right)x$ 

$34\bar{x} $ afa $x^{2}-2=2^{\dfrac {2} {3}}+2^{-\dfrac {2} {3}}$ T, T a a a a a $\left(1$ $x^{2}-2=2^{\dfrac {2} {3}}+2^{-\dfrac {2} {3}}$ then prove $that\right)$ 
$\left(1\right)$ $x=2^{\dfrac {1} {3}}+2^{-\dfrac {1} {3}}$ $0'$ $\left(11\right)$ $2x^{3}-6x=5$ $0$ 

$13.$ $x=2^{\dfrac {2} {3}}+2^{\dfrac {1} {3}}$ $x^{3}-6x=6$ 

$\left(int|imits$ $-0a1\left(1-\times n_{2}{\right)_{3}}^{n}$ 
$x^{A}3dx=7\right)$ 
$\left($ $frac\left(1\right)\left(40\right)\right)$ 
$\left($ $\left(troc\left(1\right)\left(35\right)\right)$ 
$\left(troc\left(1\right)\left(30\right)\right)$ 
$\left(tr0c\left(1\right)\left(25\right)\right)$ 

Which of the following rational numbers are 
equivalent? 
$0Ptionsy$ 
A \frac{5}{6}, \frac{30}{36} 
B $s\sqrt{rac\left(} -2\right)\left(3\right)\sqrt{1rac} \sqrt{4\right)16\right)4} $ 
C $s\sqrt{11aC\left(} -4\right)1-7b,\sqrt{1rac\left(16\sqrt{35\right)9} } $ 
D \frac{1}{2},\frac{3}{8} 

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