(9) \( x^{3}-8 \) (8) \( 2 x^{3}+x-6 x-3 \) (9) \( x^{2}+3 x \) D) \( x^{2}+3 x-10 \) (e) \( 3 a t+b b-3 a+2 b \)

It seems you have provided a list of algebraic expressions. I will simplify each expression step by step. 1. **Expression (9)**: \( x^{3}-8 \) - This is a difference of cubes, which can be factored using the formula \( a^3 - b^3 = (a - b)(a^2 + ab + b^2) \). - Here, \( a = x \) and \( b = 2 \). - Thus, we can factor it as: \[ x^{3}-8 = (x-2)(x^{2}+2x+4) \] 2. **Expression (8)**: \( 2x^{3}+x-6x-3 \) - First, combine like terms: \[ 2x^{3} + (x - 6x) - 3 = 2x^{3} - 5x - 3 \] - This expression does not factor nicely, so we will leave it as is. 3. **Expression (9)**: \( x^{2}+3x \) - Factor out the common term \( x \): \[ x^{2}+3x = x(x+3) \] 4. **Expression D**: \( x^{2}+3x-10 \) - We will factor this quadratic expression. We need two numbers that multiply to \(-10\) and add to \(3\). The numbers \(5\) and \(-2\) work. - Thus, we can factor it as: \[ x^{2}+3x-10 = (x+5)(x-2) \] 5. **Expression (e)**: \( 3at + bb - 3a + 2b \) - First, rewrite \( bb \) as \( b^2 \): \[ 3at + b^2 - 3a + 2b \] - Now, we can group the terms: \[ 3at - 3a + b^2 + 2b = 3a(t - 1) + b^2 + 2b \] - The quadratic \( b^2 + 2b \) can be factored as: \[ b^2 + 2b = b(b + 2) \] - Therefore, the entire expression can be written as: \[ 3a(t - 1) + b(b + 2) \] Now, let's summarize the simplified forms of each expression: 1. \( x^{3}-8 = (x-2)(x^{2}+2x+4) \) 2. \( 2x^{3}-5x-3 \) (not factored) 3. \( x^{2}+3x = x(x+3) \) 4. \( x^{2}+3x-10 = (x+5)(x-2) \) 5. \( 3at + b^2 - 3a + 2b = 3a(t - 1) + b(b + 2) \) If you need further assistance or specific calculations, please let me know!