If a, b, c, d are real numbers that satisfy the system of equations
abc+ab+bc+ca+a+b+c=71
bcd+bc+cd+db+b+c+d=191
cda+cd+da+ac+c+d+a=95
dab+da+ab+bd+d+a+b=143
Evaluate the value of abcd+a+b+c+d
For all four equations add 1 to each side and factor:
(a+1)*(b+1)*(c+1) = 72 = 2^3*3^2
(b+1)*(c+1)*(d+1) = 192 = 2^6*3
(a+1)*(c+1)*(d+1) = 96 = 2^5*3
(a+1)*(b+1)*(d+1) = 144 = 2^4*3^2
Now multiply everything together:
[(a+1)*(b+1)*(c+1)*(d+1)]^3 = 2^18*3^6
The cube root of each side then yields
(a+1)*(b+1)*(c+1)*(d+1) = 2^6*3^2
Take this last equation and divide it by each of the four earlier equations, in turn. Then we get
a+1 = 2^6*3^2 / 2^3*3^2 = 2^3 = 8
b+1 = 2^6*3^2 / 2^6*3 = 3^1 = 3
c+1 = 2^6*3^2 / 2^5*3 = 2*3 = 6
d+1 = 2^6*3^2 / 2^4*3^2 = 2^2 = 4
Then a=7, b=2, c=5, and d=3. Which makes abcd+a+b+c+d = 7*2*5*3+7+2+5+3 = 227.
Solution