Here we are given (x-9)^1/4 - (x+2)^1/3 = -1 and we have to find x!

As factorial is defined only for an integer, the fourth root of x - 9 should be 1 less than the third root of x + 2

We have:

1^4 = 1 , 2^4 = 16 , 3^4 = 81

1^3 = 1, 2^3 = 8, 3^3 = 27

Using these values we try to arrive at the value of x. We see that we can take x = 25 and we get x - 9 = 16 = 2^4 and x + 2 = 27 = 3^3

Also 2 - 3 = -1

**Therefore the required value of x! = 25!**

For the beggining, we'll impose constraint of existence of radical (x-9)^1/4.

x - 9 >= 0

x >= 9

So, the solutions of the equation have to be located in the range [9 ; +infinite).

NOw, to solve the equation, we'll note a = (x-9)^1/4 and b = (x+2)^1/3.

We'll raise a to the 4th power:

a^4 = x - 9 (1)

We'll raise b to the 3rd power:

b^3 = x + 2 (2)

We'll subtract (2) from (1) and we'll get:

a^4 - b^3 = x - 9 - x - 2

a^4 - b^3 = -11 (3)

a - b = -1 => b = a + 1

We'll substitute b in (3):

a^4 - (a+1)^3 = -11

a^4 - a^3 - 3a^2 - 3a - 1 = -11

a^4 - a^3 - 3a^2 - 3a + 10 = 0

We'll factorize and we'll get:

(a-2)(a^3 +a^2 - a - 5) = 0

a^3 +a^2 - a - 5 is irreducible

a - 2 = 0

a = 2

(x-9)^1/4 = 2

x - 9 = 16

x = 16+9

x = 2

25 is located in the range [9 ; infinite).

**We'll accept it as solution of the equation, so x = 25.**