Given: x - 1/x = 3√2

To find: x^3 - 1/x^3

Solution:

We can find x^3 - 1/x^3 using the identity: a^3 - b^3 = (a - b)(a^2 + ab + b^2)

Let's first square the given equation to get rid of the square root:

(x - 1/x)^2 = (3√2)^2

x^2 - 2 + 1/x^2 = 18

x^2 + 1/x^2 = 20

Now we can use the given equation to find x + 1/x:

x - 1/x = 3√2

Squaring both sides, we get:

x^2 - 2 + 1/x^2 = 18

x^2 + 1/x^2 = 20

Adding the two equations, we get:

2x^2 = 38

x^2 = 19

x = ±√19

Since x - 1/x = 3√2, we can eliminate the negative root and conclude that x = √19 + 1/√19

Now we can use this value to find x^3 - 1/x^3:

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

x - 1/x = √19 + 1/√19 - 1/(√19 + 1/√19)

Simplifying, we get:

x - 1/x = (√19 + 1/√19)^2 - 2

x - 1/x = 19 + 2√19 + 1/19 - 2

x - 1/x = 36 + 2√19

Now, let's find x^2 + x(1/x) + 1/x^2:

x^2 + x(1/x) + 1/x^2 = (√19 + 1/√19)^2 + 2

x^2 + x(1/x) + 1/x^2 = 19 + 2 + 1/19

x^2 + x(1/x) + 1/x^2 = 40 + 1/19

Multiplying the two equations, we get:

(x^3 - 1/x^3) = (36 + 2√19)(40 + 1/19)

Simplifying, we get:

(x^3 - 1/x^3) = 1440 + 72√19 + 80 + 4√19

(x^3 - 1/x^3) = 1520 + 76√19

Therefore, x^3 - 1/x^3 = 1520 + 76√19