Given the positive integers $a \le b \le c \le 100$ satisfy the system of equations below.

$\begin{cases} {a+b=c} \\ {a^4+b^2=c^2} \\ {a^3=b+c} \\ \end{cases}$

Find the sum of all possible values of $a+b+c$ .

The answer is 238.

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Solution: We see that since $a^3 = b+ c = a + b + b$ (from equation 1) that $a^3 = a + 2b$ Also, we plug in $a+b = c$ into equation 2 to get $a^4 = a^2 + 2ab$ . This is equivalent to our previous equation, so in our system of equations one equation can be discarded as it is built off the others. So we discard $a^4 + b^2 = c^2$ Now with our remaining solutions and our positive integers restriction we can plug in a=2,3,4,5 to get (a,b,c)=(2,3,5)(3,12,15)(4,30,34)(5,60,65). If a=6 then we get (6,105, 111) which is greater than 1--. So then our answer is $2+3+5+3+12+15+4+30+34+5+60+65=238$ .