X = 3 + 2√2

1/x = 1/(3 + 2√2) = 1/(3 + 2√2) × (3 - 2√2)/(3 - 2√2)

1/x = (3 - 2√2)/(9 - 8)

1/x = 3 - 2√2

√x = √(3 + 2√2)

Now, we have to solve the term √(3 + 2√2). Here, neither you can multiply it by its rationalizing factor, for can you add or subtract anything, since it's enclosed in a square root.

For solving such terms, we need to split the enclosed term {here, it is 3 + 2√2} such that it is equal to the square of any digit.

3 + 2√2 = 1 + 2 + 2√2

It is similar to the IDENTITY:

(a + b)^2 = a^2 + b^2 + 2ab

Here, a = 1

B = √2

Thus:

3 + 2√2 = 1 + 2 + 2√2 = (1 + √2)^2

Usually, the identity USED to split and compress the enclosed terms is the one used here. If not this one, the other one used is: (a - b)^2 = a^2 + b^2 - 2ab

The square root and square cancel each other. What is left is the value we need:

√x = (1 + √2) ___(1)

Now we've found √x. Similarly, we can find 1/√x

1/√x = √(1/x) = √(1/(3 + 2√2)) = √(3 - 2√2) = 1 - √2

Thus, 1/√x = (1 - √2) ____(2)

To find:

(√x - (1/√x))^3

Put (1) and (2):

=> ((1 + √2) - (1 - √2))^3

=> (1 + √2 - 1 + √2)^3

=> (√2)^3

=> 2√2