Series expansion

Question

If m and n are whole numbers and $m^{n} = 125$ then the value of $(m-5)^{n+1}=?$

Comments

m= 5 n= 3

(5)³ = 125 

now (m-5)ⁿ⁺¹ = 0

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m=125, n=1 also satisfies $m^{n}=125$

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Yes, it does. So two solutions are there.

Answer 1

125 = 5 x 5 x 5 = 5^(3) = 125 x 1 = 125^(1)

So, m=5.............. n=3
And m=125.........n=1

So, for 1st one...
(5-5)^(3+1) = 0
And...
for 2nd one...
(125-5)^(1+1) = (120)^2 = 14400