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 (x³ +y³) is ________(in integer).

The correct option is 91.

Step 1: Understand the behavior of squared terms
We are given the following equation where x and y are integers:
$${(x+y-7)}^2+{(y+3x-13)}^2=0$$
For any real numbers A and B, their squares are non-negative:
$$A^2\ge 0\text{and}{B}^2\ge 0$$
For the sum of two non-negative terms to equal zero, both individual terms must be exactly zero:
$$A^2+B^2=0\Rightarrow A=0\text{and}B=0$$

Step 2: Set up the system of linear equations
Applying this property to our equation, we obtain two simultaneous equations:
$$x+y-7=0\Rightarrow x+y=7$$ (Equation 1)
and
$$y+3x-13=0\Rightarrow 3x+y=13$$ (Equation 2)

Step 3: Solve the system of equations
We can solve for x by subtracting Equation 1 from Equation 2:
$$(3x+y)-(x+y)=13-7$$
Simplifying the terms:
$$2x=6$$
$$x=3$$
Now, substitute x = 3 back into Equation 1 to find y:
$$3+y=7$$
$$y=4$$
Since both x = 3 and y = 4 are integers, they satisfy the conditions of the problem.

Step 4: Compute the final value
We need to find the value of:
$$x^3+y^3$$
Substitute x = 3 and y = 4:
$$3^3+4^3=27+64=91$$