1.1) Simplify the following expresion fully: 1.1.1) \( (x+2)\left(x^{2}-x+3\right) \) (1.1.2) \( \frac{5}{x+3}-\frac{3}{2-x} \) 1.1.3) \( (2 y+3)\left(7 y^{2}-6 y-8\right) \) 1.1.14) \( \frac{x^{3}+1}{x^{2}+x+1} \) 1.1.5) \( \frac{3}{x^{2}-9}+\frac{2}{(x-3)^{2}} \)

For the expression \((x+2)\left(x^{2}-x+3\right)\), first, distribute \((x+2)\) across the trinomial. You get \(x(x^2) + x(-x) + x(3) + 2(x^2) + 2(-x) + 2(3)\) which simplifies to \(x^3 + 2x^2 - x + 6\). Now for \(\frac{5}{x+3}-\frac{3}{2-x}\), we can rewrite the second fraction by factoring out a negative from the denominator: \(\frac{3}{2-x} = -\frac{3}{x-2}\). The common denominator for these fractions is \((x+3)(x-2)\). Rewrite both fractions with this common denominator, combine, and simplify. Next, looking at the expression \((2 y+3)(7 y^{2}-6 y-8)\), distribute \(2y+3\) across the polynomial. After distributing, you end up with \(14y^3 - 12y^2 - 16y + 21y^2 - 18y - 24\) which simplifies down to \(14y^3 + 9y^2 - 34y - 24\). For \(\frac{x^{3}+1}{x^{2}+x+1}\), factor the numerator as \(x^3 + 1 = (x+1)(x^2-x+1)\). Notice that \(x^2+x+1\) doesn’t factor nicely over the reals, leaving your expression as \(\frac{(x+1)(x^2-x+1)}{x^2+x+1}\). This cannot be simplified further as \(x^2-x+1\) does not equal \(x^2+x+1\). Lastly, for \(\frac{3}{x^{2}-9}+\frac{2}{(x-3)^{2}}\), notice that \(x^2 - 9 = (x-3)(x+3)\). The common denominator here is \((x-3)^2(x+3)\). Rewrite each term and combine them to simplify your expression. Keep practicing, and soon these problems will feel like a breeze!