## Proving (x³ + 1/x³) = 0 given (x + 1/x)² = 3

This problem involves manipulating algebraic expressions and utilizing the given equation to arrive at the desired result. Here's how to prove it:

**1. Expand the given equation:**

(x + 1/x)² = 3 x² + 2(x)(1/x) + (1/x)² = 3 x² + 2 + 1/x² = 3

**2. Isolate the term (x² + 1/x²):**

x² + 1/x² = 3 - 2 x² + 1/x² = 1

**3. Cube the equation from step 2:**

(x² + 1/x²)³ = 1³ x⁶ + 3(x²)(1/x²) + 3(x²)(1/x⁴) + 1/x⁶ = 1 x⁶ + 3 + 3(1/x²) + 1/x⁶ = 1

**4. Simplify the equation:**

x⁶ + 1/x⁶ + 3 + 3/x² = 1 x⁶ + 1/x⁶ + 3(1 + 1/x²) = 1

**5. Substitute the value of (x² + 1/x²) from step 2:**

x⁶ + 1/x⁶ + 3(1) = 1 x⁶ + 1/x⁶ + 3 = 1

**6. Isolate the term (x⁶ + 1/x⁶):**

x⁶ + 1/x⁶ = 1 - 3 x⁶ + 1/x⁶ = -2

**7. Factor the term (x⁶ + 1/x⁶):**

(x³ + 1/x³)(x³ - 1/x³) = -2

**8. Observe that the term (x³ - 1/x³) is non-zero.**

This is because if it were zero, then (x⁶ + 1/x⁶) would also be zero, which contradicts our result in step 6.

**9. Therefore, (x³ + 1/x³) must be equal to zero.**

This is the only way for the product of two terms to be -2, while one term is non-zero.

**Therefore, we have successfully shown that (x³ + 1/x³) = 0 given (x + 1/x)² = 3.**