Calculate complex determinant $$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(a + 1)}^2}}&{{{(a + 2)}^2}}&{{{(a + 3)}^2}}\\{{b^2}}&{{{(b + 1)}^2}}&{{{(b + 2)}^2}}&{{{(b + 3)}^2}}\\{{c^2}}&{{{(c + 1)}^2}}&{{{(c + 2)}^2}}&{{{(c + 3)}^2}}\\{{d^2}}&{{{(d + 1)}^2}}&{{{(d + 2)}^2}}&{{{(d + 3)}^2}}\end{array}} \right|
$$
It's very stupid desicion for me to expand it by row or column. 
Any suggestions?
 A: Hint: Let $P(x)=x^2$. The polynomials $P(x+k)$, $k=0,1,2,3$ lies in the vector space of the polynomials of degree $\leq 2$, that is of dimension $3$. Hence there are dependant, and there exists $u,v,w\in \mathbb{R}$ such that $P(x+3)=uP(x)+vP(x+1)+wP(x+2)$ for all $x$. 
A: HINT
If you subtract the first column from every other column you get 
$$\left| {\begin{array}{*{20}{c}}{{a^2}}&{{{(2a + 1)}}}&{{{(4a + 4)}}}&{{{(6a+ 9)}}}\\{{b^2}}&{{{(2b + 1)}}}&{{{(4b + 4)}}}&{{{(6b + 9)}}}\\{{c^2}}&{{{(2c + 1)}}}&{{{(4c + 4)}}}&{{{(6c + 9)}}}\\{{d^2}}&{{{(2d + 1)}}}&{{{(4d + 4)}}}&{{{(6d + 9)}}}\end{array}} \right|
$$ 
next subtract 2 times the second column from the third column and 3 times the second column from the fourth column. Can you take it from here?
A: By the fourth column (resp. the third column, the second column) subtract the third column (resp. the second column, the first column), you get 
$\left|\begin{array}{cccc} 
a^2 & 2a+1 & 2a+3 & 2a+5 \\
b^2 & 2b+1 & 2b+3 & 2b+5 \\
c^2 & 2c+1 & 2c+3 & 2c+5 \\
d^2 & 2d+1 & 2d+3 & 2d+5 \\
\end{array} \right|=0$
