Q. If \(xy = -6\) and \(x^3 – x = y^3 – y\), find \(x^2 + y^2\).

We are given the system

\(xy = -6\)

\(x^{3} – x = y^{3} – y\)

We want to find the value of \(x^{2} + y^{2}\). The solution proceeds in clear steps.

1. Rewrite the second equation by moving terms so that one side is zero:

   \(x^{3} – y^{3} = x – y\)

   Factor the left-hand side using the difference of cubes formula and factor the right-hand side as well:

   \((x – y)\bigl(x^{2} + xy + y^{2}\bigr) = (x – y)\)

   Bring all terms to one side (factor out the common factor \((x-y)\)) to obtain:

   \((x – y)\bigl(x^{2} + xy + y^{2} – 1\bigr) = 0\)

2. From the product above, at least one factor must be zero. Thus there are two cases to consider:

   1. Case 1: \(x – y = 0\), i.e. \(x = y\).

      Substitute into \(xy = -6\): \(x^{2} = -6\). Then

      \(x^{2} + y^{2} = x^{2} + x^{2} = 2x^{2} = 2(-6) = -12\).

      Note: this value corresponds to complex solutions for \(x\) and \(y\) (since \(x^{2} = -6\) implies \(x\) is not real). If we are restricting to real numbers, this case is not allowed.

   2. Case 2: \(x^{2} + xy + y^{2} – 1 = 0\), i.e.

      \(x^{2} + xy + y^{2} = 1\).

      Use the given \(xy = -6\) and substitute:

      \(x^{2} + (-6) + y^{2} = 1\)

      So

      \(x^{2} + y^{2} – 6 = 1\)

      and therefore

      \(x^{2} + y^{2} = 7\).

      This case yields real solutions (since it does not force a negative square), so for real \(x,y\) this is the valid value.

3. Conclusion:

   If \(x\) and \(y\) are allowed to be complex and \(x = y\) occurs, then \(x^{2} + y^{2} = -12\). If we restrict to real numbers (the usual interpretation), the admissible case is the second one, giving

   \(x^{2} + y^{2} = 7\).