1. **State the problem:** Solve the equation $$\left(a+\sqrt{2}\right)^2 + \left(\sqrt{2}a - 1\right)^2 = 6$$ for the variable $a$. 2. **Expand each squared term:** $$\left(a+\sqrt{2}\right)^2 = a^2 + 2\sqrt{2}a + 2$$ $$\left(\sqrt{2}a - 1\right)^2 = 2a^2 - 2\sqrt{2}a + 1$$ 3. **Substitute expansions back into the equation:** $$a^2 + 2\sqrt{2}a + 2 + 2a^2 - 2\sqrt{2}a + 1 = 6$$ 4. **Combine like terms:** The $2\sqrt{2}a$ and $-2\sqrt{2}a$ cancel out: $$a^2 + 2a^2 + 2 + 1 = 6$$ $$3a^2 + 3 = 6$$ 5. **Isolate the quadratic term:** $$3a^2 = 6 - 3$$ $$3a^2 = 3$$ 6. **Divide both sides by 3:** $$\cancel{3}a^2 = \cancel{3}$$ $$a^2 = 1$$ 7. **Solve for $a$ by taking the square root:** $$a = \pm \sqrt{1}$$ $$a = \pm 1$$ **Final answer:** $$a = 1 \text{ or } a = -1$$