$\begin{cases} a^3 + 12ab+ 64=b^3 \\ a^4 + 256=b^4 \end{cases}$

If $(a_1,b_1),(a_2,b_2) ,\ldots , (a_n,b_n)$ are real numbers which satisfy the system of equations above, find the value of

$\displaystyle 1 + 27 \sum_{j=1}^n (b_j^3 - a_j^3).$

The answer is 1729.

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$a^3 - b^3 + 64 + 12ab = 0$

$\dfrac{1}{2}(a - b + 4)((a+b)^2 + (b + 4)^2 + (4-a)^2) = 0$

So there are two possible casesCASE1 When $(a+b)^2 + (b + 4)^2 + (4-a)^2 = 0$

Then $a = 4, b = -4$

But it doesnt satisfy

CASE2 When $a - b + 4 = 0$ that is $a+4=b$

Putting this value in the second equation :

$a^4 + 256 = (a+4)^4$

Solving it, we get a cubic equation, we can found the only possible pair is $(a,b)=(0,4)$

So our answer is 1729