Note that your formula as submitted is incorrect: the 4 belongs in the numerator of the expression, not in the exponent.
The statement is trivially true for n=1
Assume it is true for n
Add (n+1)3 to the formula for n to get
[(n4 + 2n3 + n2) + 4(n3 + 3n2 + 3n +1)]/4
Expand and collect the terms to get (n4 + 6n3 + 13n2 + 12n + 4)/4
Divide the numerator by n2+ 2n + 1 to get n2 + 4n + 4
Therefore the sum to the term in n+1 is (n+1)2 (n+2)2/4 which is the formula with n+1 substituted for n.
Q.E.D. by induction!