If $(a+1) (b+1) (c+1) (d+1)=3,$$(a+2) (b+2) (c+2) (d+2)=7,$$(a+3) (b+3) (c+3) (d+3)=13,$$(a+4) (b+4) (c+4) (d+4)=21$. Then the value of $(a+5) (b+5) (c+5) (d+5)$ is

Lets consider a function $f\left(x\right)=(x+a)(x+b)(x+c)(x+d)-(x^2+x+1)$  
which is a monic quartic function
Then we can observe that  $f\left(1\right)=f\left(2\right)=f\left(3\right)=f\left(4\right)=0$
So we can write $f\left(x\right)$  as $(x-1)(x-2)(x-3)(x-4)$  because
$f\left(x\right)$  is a monic quartic function 
So,  $f\left(x\right)=(x+a)(x+b)(x+c)(x+d)-(x^2+x+1)$
$=(x-1)(x-2)(x-3)(x-4)$ 
Now put x = 5 we get
$(a+5)(b+5)(c+5)(d+5)$
$=4\times 3\times 2\times 1+(5^2+5+1)=55$