Question

\( (y + 3x - 13)^3 + (x + y - 7)^2 = 0 \) where x and y are integers. The value of \(x^3 + y^3\) is _____.

Hint:

When faced with an equation of the form \(f(x,y)^a + g(x,y)^b = 0\) with integer solutions, especially if one exponent is even, first check the simple case where both \(f(x,y)=0\) and \(g(x,y)=0\). This often leads to the intended unique integer solution.

Answer 1

Step 1: Analyze equation structure.
Let $A = y + 3x - 13$ and $B = x + y - 7$. $A^3 + B^2 = 0$. Since $B^2 \geq 0$ for all real $B$, then $A^3$ must be $\leq 0 \implies A \leq 0$. For typical integer solution problems of this form, the most direct solution is $A=0$ and $B=0$.
Step 2: Solve the Linear System.
1) $3x + y = 13$ 2) $x + y = 7$ Subtracting (2) from (1): $2x = 6 \implies x = 3$. Substituting $x=3$ in (2): $3 + y = 7 \implies y = 4$.
Step 3: Calculate requested value.
$x^3 + y^3 = 3^3 + 4^3 = 27 + 64 = 91$.