數學 M2 mathematic induction

a)It is true for n = 1 since 1³ + 2³ + 3³ = 36 is divisible by 9.
If P(k) is assumed to be true for n = k ,
i.e. k³ + (k+1)³ + (k+2)³ = 9m , were m is a positive integer.
When n = k+1 ,
(k+1)³ + (k+2)³ + (k+3)³
= (k+1)³ + (k+2)³ + k³ + 9k² + 27k + 27
= 9m + 9k² + 27k + 27
= 9(m + k² + 3k + 3) is divisible by 9.
(Proved by MI)

b)The total volume of the three consecutive even cubes
= (2a)³ + (2(a+1))³ + (2(a+2))³
= 2³a³ + 2³(a+1)³ + 2³(a+2)³
= 8 (a³ + (a+1)³ + (a+2)³)
By the result of a) , we let a³ + (a+1)³ + (a+2)³ = 9m were m is a positive integer, then it becomes8 (9m)
= 72m is a multipe of 72.

c)
If the three cubes are recast to form a larger cube , the volume
= 72m
= 2³ * 3² m
Then m = 3n³ were n is a positive integer for 72m is a perfect cube.
When n = 1 ,
i.e. m = 3 ,
a³ + (a+1)³ + (a+2)³ = 9m = 27 have no positive integer solutions.
When n = 2 ,
i.e. m = 3 * 2³ = 24 ,
a³ + (a+1)³ + (a+2)³ = 9m = 216 ,
a = 3
∴ The least possible value of a = 3 , when the total volume of the three cubes
= 6³ + 8³ + 10³ = 12³ .

Q0Q唔知我有冇投錯呢~!~

a)
assume P(k) is true, ie, k³ + (k+1)³ + (k+2)³ = 9m for some positive integer m,
for n=k+1,
(k+1)³ + (k+2)³ + (k+3)³
= (k+1)³ + (k+2)³ + (k³+ 9k²+27k+ 27)
= k³ + (k+1)³ + (k+2)³ + 9(k²+3k+3)
= 9(m+k²+3k+3)
so P(k+1) is true.
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b)
the length of the sides of the cubes are 2a, 2a+2 and 2a+4 respectively

sum of their volume
= (2a)³ + (2a+2)³ + (2a+4)³
= 8[a³+(a+1)³+(a+2)³]
= 72k, which is a multipe of 72
since a³+(a+1)³+(a+2)³ = 9k for some positive integer k, (by (a))

i am so sorry that i cannot understand the question "find the least possible value of a". i suppose the answer should be 1 ?
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