Recall the binomial theorem:
[tex](a+b)^n=\displaystyle\sum_{k=0}^n\binom nk a^k b^{n-k}[/tex]
where
[tex]\dbinom nk=\dfrac{n!}{k!(n-k)!}[/tex]
is the binomial coefficient. So
[tex](x+1)^n=\displaystyle\sum_{k=0}^n\binom nk x^k[/tex]
The x² term occurs for k = 2, which has coefficient
[tex]\dbinom n2=\dfrac{n!}{2!(n-2)!}=\dfrac{n(n-1)}2[/tex]
The x³ occurs for k = 3, with coefficient
[tex]\dbinom n3=\dfrac{n!}{3!(n-3)!}=\dfrac{n(n-1)(n-2)}6[/tex]
The latter coefficient is twice the other one, so that
[tex]\dfrac{n(n-1)(n-2)}6=2\cdot\dfrac{n(n-1)}2[/tex]
Solve for n :
[tex]\dfrac{n(n-1)(n-2)}6=n(n-1)[/tex]
[tex]n(n-1)(n-2)=6n(n-1)[/tex]
[tex]n-2=6[/tex]
[tex]\boxed{n=8}[/tex]
