If x⁴ + 1/x⁴ = 47, where x > 0, then the value of x³ + 1/x³ is

Squaring a sum of the form (a + 1/a) always gives (a + 1/a)² = a² + 2 + 1/a²; subtracting 2 and taking a square root therefore steps a higher power sum down to a lower one, letting x⁴ + 1/x⁴ lead to x² + 1/x² and then to x + 1/x. The cube sum follows from x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x), which expands the cube and removes the extra linear terms.

1. Add 2 to both sides of the given equation: (x² + 1/x²)² = x⁴ + 2 + 1/x⁴ = 47 + 2 = 49.

2. Since x² + 1/x² is at least 2 for every real nonzero x, take the positive square root: x² + 1/x² = 7.

3. Repeat the same step: (x + 1/x)² = x² + 2 + 1/x² = 7 + 2 = 9. Since x > 0, x + 1/x is also positive, so take the positive square root: x + 1/x = 3.

4. Apply the cube-sum identity x³ + 1/x³ = (x + 1/x)³ − 3(x + 1/x): substitute x + 1/x = 3 to get 3³ − 3×3 = 27 − 9 = 18.

An equivalent identity, x³ + 1/x³ = (x + 1/x)(x² − 1 + 1/x²), gives the same result: 3 × (7 − 1) = 3 × 6 = 18, confirming the value obtained above.

So x³ + 1/x³ = 18.