Let R b e a ommutative ring of hara teristi 3 and with unity and let a, b, c ∈R. Compute and simplify (a + b)3; hen e ompute and simplify (a + b)9

It seems like there are some typos in your question. I believe you're asking for the simplification of (a + b)³ and (a + b)⁹ in a commutative ring of characteristic 3. Here's how you can do it:

1. (a + b)³:

Using the binomial theorem, we have (a + b)³ = a³ + 3a²b + 3ab² + b³.

However, since we are in a ring of characteristic 3, 3 is equivalent to 0 in this ring. Therefore, 3a²b and 3ab² both become 0.

So, (a + b)³ simplifies to a³ + b³.

2. (a + b)⁹:

Again, using the binomial theorem, we have (a + b)⁹ = a⁹ + 9a⁸b + 36a⁷b² + 84a⁶b³ + 126a⁵b⁴ + 126a⁴b⁵ + 84a³b⁶ + 36a²b⁷ + 9ab⁸ + b⁹.

However, since we are in a ring of characteristic 3, any term with a coefficient that is a multiple of 3 is equivalent to 0 in this ring. Therefore, all terms except a⁹ and b⁹ become 0.

So, (a + b)⁹ simplifies to a⁹ + b⁹.