
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.