# Difficult equation without unknown numbers

Algebra

### Difficult equation without unknown numbers

The following Tn formulas calculate the n-th term of the Tribonacci sequence. The (2) is unknown and uses only one cubic root, since we can write $$b = (p + (4/p) + 1)/3$$, but how it is proved that (1) = (2)?

$$p = \sqrt [3] {19 + 3 \sqrt {33}}, p' = \sqrt [3] {19 - 3 \sqrt {33}}$$

$$q = \sqrt [3] {586 + 102 \sqrt {33}}, q' = \sqrt [3] {586 - 102 \sqrt {33}}$$

$$pp' = qq' = 4$$

$$b = (p + p' + 1)/ 3, d = (q + q' + 1)/ 3$$

$$T_n = [ b^{n - 1}/ (d - 1)] (1)$$

$$T_n = [(b - 1)b^n/(4b - 6)] (2)$$
Guest

### Re: Difficult equation without unknown terms

Guest wrote:The following Tn formulas calculate the n-th term of the Tribonacci sequence. The (2) is unknown and uses only one cubic root, since we can write $$b = (p + (4/p) + 1)/3$$, but how it is proved that (1) = (2)?

$$p = \sqrt [3] {19 + 3 \sqrt {33}}, p' = \sqrt [3] {19 - 3 \sqrt {33}}$$

$$q = \sqrt [3] {586 + 102 \sqrt {33}}, q' = \sqrt [3] {586 - 102 \sqrt {33}}$$

$$pp' = qq' = 4$$

$$b = (p + p' + 1)/ 3, d = (q + q' + 1)/ 3$$

$$T_n = [ b^{n + 1}/ (d - 1)] (1)$$

$$T_n = [(b - 1)b^n/(4b - 6)] (2)$$
Guest