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

Question

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

Asked on March 26, 2026Mathematics

This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.

Accepted Answer · 2026-03-26T05:31:27.941Z

Step 1: Find four rational numbers between 3 and 4. 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 $\boxed{\frac{16}{5}, \frac{17}{5}, \frac{18}{5}, \frac{19}{5}}$. Step 2: Find the value of $a$ for which $(x+2a)$ is a factor of $x^5 - 4a^2x^3 + 2x + 2a + 3$. 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 $\boxed{\frac{3}{2}}$. Step 3: Draw a histogram to represent the data. 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. Step 4: Factorise the following using suitable identities. 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 $\boxed{(\sqrt{3}a + 2b - 3c)(3a^2 + 4b^2 + 9c^2 - 2\sqrt{3}ab + 6bc + 3\sqrt{3}ac)}$.

Accepted Answer · 2026-03-26T05:31:27.941Z

Step 1: Find four rational numbers between 3 and 4. 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 $\boxed{\frac{16}{5}, \frac{17}{5}, \frac{18}{5}, \frac{19}{5}}$. Step 2: Find the value of $a$ for which $(x+2a)$ is a factor of $x^5 - 4a^2x^3 + 2x + 2a + 3$. 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 $\boxed{\frac{3}{2}}$. Step 3: Draw a histogram to represent the data. 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. Step 4: Factorise the following using suitable identities. 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 $\boxed{(\sqrt{3}a + 2b - 3c)(3a^2 + 4b^2 + 9c^2 - 2\sqrt{3}ab + 6bc + 3\sqrt{3}ac)}$.