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: Reduce the equation to two linear conditions.
Because $(x+y-7)^2 \geq 0$ and $(y+3x-13)^2 \geq 0$ for any real $x,y$, their sum can equal zero only if both terms are zero individually:
$$x+y-7=0 \quad \text{and} \quad y+3x-13=0$$

Step 2: Solve for $x+y$ and $x$ using elimination instead of substitution.
Write the two equations as:
$$x+y=7 \qquad \text{...(i)}$$
$$3x+y=13 \qquad \text{...(ii)}$$
Subtract (i) from (ii):
$$(3x+y)-(x+y)=13-7$$
$$2x = 6 \implies x = 3$$
Then from (i): $y = 7 - 3 = 4$.

Step 3: Use the sum-of-cubes identity instead of computing cubes directly.
Recall the identity:
$$x^3+y^3=(x+y)(x^2-xy+y^2)$$
We already know $x+y=7$. Now find $xy$ and $x^2+y^2$:
$$xy = 3\times4 = 12$$
$$x^2+y^2 = (x+y)^2 - 2xy = 7^2 - 2(12) = 49-24=25$$
So:
$$x^2-xy+y^2 = 25 - 12 = 13$$

Step 4: Combine using the identity.
$$x^3+y^3 = (x+y)(x^2-xy+y^2) = 7\times 13 = 91$$
$$\boxed{x^3+y^3=91}$$