System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$

Note that \begin{align} f(k)&=(a+k)(b+k)(c+k)(d+k) \\ &=k^4+(a+b+c+d)k^3+(a b+a c+a d+b c+b d+c d) k^2 +(a b c+a b d+a c d+b c d) k+a b cd \\ &=k^4 + k^3 e_1 + k^2 e_2 + k e_3 + e_4. \end{align}

Now you have a linear system of 4 equations and 4 unknowns ($e_1,e_2,e_3,e_4$). Solve it and find $f(5)$.

First, by expanding the four equations, we note that the system of equations is equivalent to $$ \begin{bmatrix} 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ 1 & 2 & 2 & 4 & 2 & 4 & 4 & 8 & 2 & 4 & 4 & 8 & 4 & 8 & 8 \\ 1 & 3 & 3 & 9 & 3 & 9 & 9 & 27 & 3 & 9 & 9 & 27 & 9 & 27 & 27 \\ 1 & 4 & 4 & 16 & 4 & 16 & 16 & 64 & 4 & 16 & 16 & 64 & 16 & 64 & 64 \end{bmatrix} \begin{bmatrix} abcd \\ abc \\ abd \\ ab \\ acd \\ ac \\ ad \\ a \\ bcd \\ bc \\ bd \\ b \\ cd \\ c \\ d \end{bmatrix} = \begin{bmatrix} 14 \\ 29 \\ 52 \\ 83 \end{bmatrix}. $$ Using row reduction, we see that this is equivalent to the matrix equation $$ \begin{bmatrix} 1 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 & 0 & 1 & 1 & 0 & 0 & 1 & 1 & 0 & 1 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 1 & 0 & 1 & 1 \end{bmatrix} \begin{bmatrix} abcd \\ abc \\ abd \\ ab \\ acd \\ ac \\ ad \\ a \\ bcd \\ bc \\ bd \\ b \\ cd \\ c \\ d \end{bmatrix} = \begin{bmatrix} 7 \\ 3 \\ 4 \\ 0 \end{bmatrix}. $$ This resolves into the following system of equations: $$ \begin{aligned} abcd = 7 \\ abc + abd + acd + bcd = 3 \\ ab + ac + ad + bc + bd + cd = 4 \\ a + b + c + d = 0. \end{aligned} $$

Finally, we note that $(a+5)(b+5)(c+5)(d+5)$ is equal to $$abcd + 5(abc + abd + acd + bcd) + 25(ab + ac + ad + bc + bd + cd) + 125(a + b + c + d) + 625.$$ So $(a+5)(b+5)(c+5)(d+5) = 7 + (5\times3) + (25\times4) + (125\times0) + (625) = 747$.