Factorise the following: 3√3 a³ + 8b³ - 27c³ + 18√3b abc

To find rational numbers between 3 and 4, we can express them as fractions with a common denominator. Let's use a denominator of 5.

$$3 = \frac{3 \times 5}{5} = \frac{15}{5}$$ $$4 = \frac{4 \times 5}{5} = \frac{20}{5}$$

The rational numbers between $\frac{15}{5}$ and $\frac{20}{5}$ are $\frac{16}{5}, \frac{17}{5}, \frac{18}{5}, \frac{19}{5}$.

The four rational numbers are $\boxed{\frac{16}{5}, \frac{17}{5}, \frac{18}{5}, \frac{19}{5}}$.

According to the Factor Theorem, if $(x+2a)$ is a factor of a polynomial $P(x)$, then $P(-2a) = 0$.
Let $P(x) = x^5 - 4a^2x^3 + 2x + 2a + 3$.
Substitute $x = -2a$ into $P(x)$:

$$P(-2a) = (-2a)^5 - 4a^2(-2a)^3 + 2(-2a) + 2a + 3$$ $$P(-2a) = -32a^5 - 4a^2(-8a^3) - 4a + 2a + 3$$ $$P(-2a) = -32a^5 + 32a^5 - 2a + 3$$ $$P(-2a) = -2a + 3$$

Set $P(-2a) = 0$:

$$-2a + 3 = 0$$ $$-2a = -3$$ $$a = \frac{-3}{-2}$$ $$a = \frac{3}{2}$$

The value of $a$ is $\boxed{\frac{3}{2}}$.

To draw a histogram for the given data:

• Draw two axes. Label the horizontal axis "Age (in years)" and the vertical axis "Number of children (Frequency)".
• Mark the class boundaries on the horizontal axis: 10, 20, 30, 40, 50, 60, 70, 80.
• Choose a suitable scale for the vertical axis. Since the maximum frequency is 12, a scale where 1 unit represents 1 child would be appropriate.
• Draw rectangular bars for each class interval. The width of each bar will correspond to the class interval (e.g., from 10 to 20, 20 to 30, etc.), and the height of each bar will correspond to its frequency. The bars should be adjacent to each other without any gaps.

Here is a description of the bars:

• For age group 10-20, draw a bar of height 4.
• For age group 20-30, draw a bar of height 3.
• For age group 30-40, draw a bar of height 6.
• For age group 40-50, draw a bar of height 12.
• For age group 50-60, draw a bar of height 9.
• For age group 60-70, draw a bar of height 10.
• For age group 70-80, draw a bar of height 4.

Option 1: Factorise $x^2 - y^2 - 2x + 1$.
Rearrange the terms to group the $x$ terms:

$$x^2 - 2x + 1 - y^2$$

Recognize that $x^2 - 2x + 1$ is a perfect square trinomial, which can be written as $(x-1)^2$.

$$(x-1)^2 - y^2$$

This expression is in the form of a difference of squares, $A^2 - B^2 = (A-B)(A+B)$, where $A = (x-1)$ and $B = y$.

$$((x-1) - y)((x-1) + y)$$ $$(x - y - 1)(x + y - 1)$$

The factorised expression is $\boxed{(x - y - 1)(x + y - 1)}$.

Option 2: Factorise $3\sqrt{3}a^3 + 8b^3 - 27c^3 + 18\sqrt{3}abc$.
This expression resembles the identity $X^3 + Y^3 + Z^3 - 3XYZ = (X+Y+Z)(X^2+Y^2+Z^2-XY-YZ-ZX)$.
Let's identify $X, Y, Z$:

$$X^3 = 3\sqrt{3}a^3 = (\sqrt{3}a)^3 \implies X = \sqrt{3}a$$ $$Y^3 = 8b^3 = (2b)^3 \implies Y = 2b$$ $$Z^3 = -27c^3 = (-3c)^3 \implies Z = -3c$$

Now, check the term $-3XYZ$:

$$-3XYZ = -3(\sqrt{3}a)(2b)(-3c)$$ $$-3XYZ = -3 \times \sqrt{3} \times 2 \times (-3) \times abc$$ $$-3XYZ = 18\sqrt{3}abc$$

This matches the last term in the given expression.
So, we can apply the identity:

$$(\sqrt{3}a + 2b - 3c)((\sqrt{3}a)^2 + (2b)^2 + (-3c)^2 - (\sqrt{3}a)(2b) - (2b)(-3c) - (-3c)(\sqrt{3}a))$$ $$(\sqrt{3}a + 2b - 3c)(3a^2 + 4b^2 + 9c^2 - 2\sqrt{3}ab + 6bc + 3\sqrt{3}ac)$$

The factorised expression is $\boxed{(\sqrt{3}a + 2b - 3c)(3a^2 + 4b^2 + 9c^2 - 2\sqrt{3}ab + 6bc + 3\sqrt{3}ac)}$.