Question

The value of (n!)^n if n+(n-1)+(n-2)=n (n-1)(n-2)where n^3>9 a is positive number

Answer 1

n + (n-1) + (n-2) = n(n-1)(n-2)
⇒ 3n-3 = n(n-1)(n-2)
⇒ 3 (n-1) = n(n-1)(n-2)
⇒n(n-1)(n-2) - 3(n-1) = 0
⇒(n-1)(n²-2n-3) = 0
⇒(n-1)(n-3)(n+1) = 0
⇒ n = 1, 3 , -1

now condition given n³ > 9 ,

       is only satisfied by n = 3 [ 1³ = 1 < 9 , -1³ = -1 <9]

so n = 3

n!ⁿ = 3!³ = 6³ = 216

Hope this helps.

Answer 2

Answer:

6^3=216

Step-by-step explanation:

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