Let's break down and solve the exercises from the image you uploaded:

Factorizations:

1. A(x) = x³ - 8
   This is a difference of cubes:
   $x^3 - 8 = (x - 2)(x^2 + 2x + 4)$
2. B(x) = x³ - 8 - 4(6x² - 4) + 5x - 10
   Simplify the expression:
   $B(x) = x^3 - 8 - 24x^2 + 16 + 5x - 10$ $B(x) = x^3 - 24x^2 + 5x - 2$
3. C(x) = x³ + 27
   This is a sum of cubes:
   $x^3 + 27 = (x + 3)(x^2 - 3x + 9)$
4. D(x) = x³ + 27 + 3(x² - 9) + x - 3
   Simplify:
   $D(x) = x^3 + 27 + 3x^2 - 27 + x - 3$ $D(x) = x^3 + 3x^2 + x - 3$

1) a) Calculations:

• A:
  Expression to simplify:
  $A = \left( \frac{b}{2} \right)^{-2} - \frac{\sqrt{5}}{3\sqrt{5}} \cdot [-1]$ The first part:
  $\left( \frac{b}{2} \right)^{-2} = \left( \frac{2}{b} \right)^2 = \frac{4}{b^2}$ The second part simplifies to:
  $\frac{\sqrt{5}}{3\sqrt{5}} = \frac{1}{3}$ Thus,
  $A = \frac{4}{b^2} - \left(-\frac{1}{3}\right) = \frac{4}{b^2} + \frac{1}{3}$

• B:
  Expression:
  $B = \frac{\Lambda}{3\sqrt{5}} - \frac{\Lambda}{5 + \sqrt{5}}$

1) b) Show that:

$\frac{3}{\sqrt{5}} + \frac{12}{5 + \sqrt{5}} \in \mathbb{N}$ First, rationalize both terms. For the first term:
$\frac{3}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \frac{3\sqrt{5}}{5}$ For the second term:
$\frac{12}{5 + \sqrt{5}} \cdot \frac{5 - \sqrt{5}}{5 - \sqrt{5}} = \frac{12(5 - \sqrt{5})}{(5 + \sqrt{5})(5 - \sqrt{5})} = \frac{12(5 - \sqrt{5})}{25 - 5} = \frac{12(5 - \sqrt{5})}{20} = \frac{3(5 - \sqrt{5})}{5}$ Adding both terms:
$\frac{3\sqrt{5}}{5} + \frac{3(5 - \sqrt{5})}{5} = \frac{3\sqrt{5}}{5} + \frac{15 - 3\sqrt{5}}{5} = \frac{15}{5} = 3$ Hence, $3 \in \mathbb{N}$, so the result holds.

2) Develop and simplify $(2 \cdot \sqrt{3})^4$:

$(2\sqrt{3})^4 = (2^4)(\sqrt{3}^4) = 16 \cdot 9 = 144$

Now, simplify $(2 + \sqrt{3})^2$: $(2 + \sqrt{3})^2 = 2^2 + 2(2)(\sqrt{3}) + (\sqrt{3})^2 = 4 + 4\sqrt{3} + 3 = 7 + 4\sqrt{3}$

3) Factorize $\Phi = 3x(x - 1) - (x^2 - 1)$:

Expand:
$\Phi = 3x(x - 1) - (x^2 - 1) = 3x^2 - 3x - x^2 + 1$ Simplify:
$\Phi = 2x^2 - 3x + 1$ Now, factor the quadratic:
$\Phi = (2x - 1)(x - 1)$

4) Given that $a$ and $b$ are non-zero real numbers, prove that:

$\left( \frac{a}{b} \right)^n b^{n + 1} a^{-n} = b$ The left-hand side simplifies to:
$\left( \frac{a}{b} \right)^n b^{n + 1} a^{-n} = \frac{a^n}{b^n} \cdot b^{n + 1} \cdot \frac{1}{a^n} = \frac{b^{n + 1}}{b^n} = b$ Hence, the equation is true.

Would you like more details on any step? Here are five related questions to practice:

1. How can you factor higher powers of polynomials using special formulas?
2. What are the steps to rationalize fractions with square roots in the denominator?
3. Can you prove the identity for simplifying $(a/b)^n b^{n+1} a^{-n}$?
4. How would you apply the distributive property to expand binomials like $(x + y)^n$?
5. What methods can you use to factorize polynomials of higher degrees?

Tip: Always simplify square root expressions before proceeding with operations, such as addition or multiplication!