Question

$\sqrt[3]{ \sqrt[3]{2} - 1} = \sqrt[3]{a} + \sqrt[3]{b} + \sqrt[3]{ \frac{1}{9} } \\ \\ If \: a>b \: \: , \: \: Find \: \: (a+2b) \:.$

Answer 1

Step One
Subtract cube root 1/9 to the left hand side. Or subtract cube root (1/9) from both sides.
$\sqrt[3]{ \sqrt[3]{2} -1 } - \sqrt[3]{ \frac{1}{9} } = \sqrt[3]{a} + \sqrt[3]{b}$

Step Two. 
There is a minus sign in front of ${-}\sqrt[3]{ \frac{1}{9} }$
We must get rid of it. Because it is a minus in front of a cube root, we can bring it inside the cube root sign like so, and make it a plus out side the cube root sign
 ${+}\sqrt[3]{ \frac{-1}{9} }$
 
Step Three
Write the Left side with the minus sign placed in the proper place
$\sqrt[3]{ \sqrt[3]{2} -1 } + \sqrt[3]{ \frac{-1}{9} } = \sqrt[3]{a} + \sqrt[3]{b}$

Step Four
Equate cube root b with cube root (-1/9)
$\sqrt[3]{b} = \sqrt[3]{ \frac{-1}{9} }$

Step Five
Equate the cube root of a with what's left over on the left
$\sqrt[3]{ \sqrt[3]{2} -1 } = \sqrt[3]{a}$

Step 6. 
I'll just work with b for a moment.
Cube both sides of  cube root (b) = cube root (-1/9)
$\sqrt[3]{b} ^{3} =\sqrt[3]{ \frac{-1}{9} }^3}$
$\text{b =} \frac{-1}{9}$
$\text{2b =}\frac{-2}{9}$

Step seven
the other part is done exactly the same way
a = cuberoot(2) - 1.

What you do from here is up to you. It is not pleasant.
Is this clearer?

a + 2b should come to cuberoot(2) - 1 - 2/9
a + 2b should come to cuberoot(2) - 11/9

I hope a person is marking this. I wonder how many of your class mates got it. 

Answer 2

Define
$c=\sqrt[3]{\sqrt[3]{2}-1}-\sqrt[3]{\frac{1}{9}} \approx 0.157435964092$

Then you have the symmetrical equation
$c=\sqrt[3]{a}+\sqrt[3]{b}$
which can be solved for b to give
$b=(c-\sqrt[3]{a})^{3}$

Substituting into your expression gives
$a+2b=a+2(c-\sqrt[3]{a})^{3}$

The requirement that a > b means this is only relevant for
$a > (\frac{c}{2})^{3} \approx 0.000487777605001$

The attached graphs show the general shape of a+2b and some detail near the origin. "a" is plotted on the x-axis; "b" is plotted on the y-axis.