What will be the value of the following determinant without expanding it?

If you expand the binomials and you subtract the $1^{\rm st}$ column from the 2nd, 3rd, and 4th you get $$ \begin{vmatrix}a^2 & 2a+1 & 4a+4 & 6a+9 \\ b^2 & 2b+1 & 4b+4 & 6b+9 \\ c^2 & 2c+1 & 4c+4 & 6c+9 \\ d^2 & 2d+1 & 4d+4 & 6d+9\end{vmatrix}. $$ Next subtract 2 times the 2$^{\rm nd}$ column from the 3$^{\rm rd}$ one, and 3 times the 2$^{\rm nd}$ column from the 4$^{\rm th}$ one: $$ \begin{vmatrix}a^2 & 2a+1 & 2 & 6 \\ b^2 & 2b+1 & 2 & 6 \\ c^2 & 2c+1 & 2 & 6 \\ d^2 & 2d+1 & 2 & 6\end{vmatrix}=0. $$ Now it is clear that the determinant is zero, as the fourth column is a multiple of the third one (or, factor 3 from the fourth column to get two equal columns).

Hint. Is it possible to find $A,B,C$ such that for all $x\in\mathbb{R}$, $$(x+3)^2=A(x+2)^2+B(x+1)^2+Cx^2?$$

P.S. The answer is yes: $A = 3$, $B = -3$, $C = 1$.

Switch the sign of columns 2. and 4., then multiply column 2. and 3. by 3 and finally add them up to get zero in every row:

Col.1 - 3 Col.2 + 3 Col.3 - Col.4 = 0

So columns are linearly dependent, hence the determinant is zero for any $a\ldots d$.