Question

It is given that \(x\) and \(y\) are integers in the following equation:
\[ (x+y-7)^2+(y+3x-13)^2=0 \]
The value of \((x^3+y^3)\) is ________ (in integer).

Hint:

A sum of two real squares is zero only if each square is zero; solve the resulting linear pair for \(x\), \(y\).

Answer 1

Correct Answer: 91

Step 1: Understanding the Question.
We are given that the sum of two squared real quantities equals zero, and \(x\), \(y\) are integers. We need to find \(x^3+y^3\).

Step 2: Key Formula or Approach.
A sum of two real squares can only be zero if each square is individually zero, since a real square can never be negative. So:
\[ x+y-7 = 0 \quad \text{and} \quad y+3x-13 = 0 \]
This converts the single equation into two simultaneous linear equations in \(x\) and \(y\).

Step 3: Detailed Explanation.
From the first equation:
\[ x + y = 7 \quad \Rightarrow \quad y = 7 - x \]
Substitute this into the second equation:
\[ (7-x) + 3x - 13 = 0 \]
\[ 7 - x + 3x - 13 = 0 \]
\[ 2x - 6 = 0 \]
\[ x = 3 \]
Then:
\[ y = 7 - 3 = 4 \]
Check both original conditions: \(x+y-7 = 3+4-7=0\), and \(y+3x-13 = 4+9-13=0\). Both are satisfied, and both \(x=3\), \(y=4\) are integers, as required.

Step 4: Compute \(x^3+y^3\).
\[ x^3 = 3^3 = 27, \qquad y^3 = 4^3 = 64 \]
\[ x^3+y^3 = 27+64 = 91 \]

Step 5: Final Answer.
\[ \boxed{x^3+y^3 = 91} \]