Answer
**QUESTION 1:**
1.1.1 \( x^2 - y^2 = (x - y)(x + y) \)
1.1.2 \( 625a^4 - 81b^8 = (25a^2 - 9b^4)(25a^2 + 9b^4) \)
1.2 General Statement: Any expression of the form \( A^2 - B^2 \) can be factorised as \( (A - B)(A + B) \).
**QUESTION 2:**
2.1.1 \( x^2 + 2xy + y^2 = (x + y)^2 \)
2.1.2 \( x^2 - 8x + 16 = (x - 4)^2 \)
2.1.3 \( 16a^4 - 24a^2b^3 + 9b^6 = (4a^2 - 3b^3)^2 \)
2.2 General Statement: A perfect square trinomial of the form \( A^2 \pm 2AB + B^2 \) can be factorised as \( (A \pm B)^2 \).
**QUESTION 3:**
3.1.1 \( x^3 + y^3 = (x + y)(x^2 - xy + y^2) \)
3.1.2 \( 8a^6 + 729 = (2a^2 + 9)(4a^4 - 18a^2 + 81) \)
3.1.3 \( (3x + y)^3 + (3x - y)^3 = 18x(3x^2 + y^2) \)
Solution
### QUESTION 1
#### 1.1 Factorise fully and simplify where possible
**1.1.1**
We have
\[
x^2 - y^2
\]
This is a difference of two squares, so it can be factorised as
\[
(x-y)(x+y).
\]
**1.1.2**
We have
\[
625a^4-81b^8.
\]
Notice that
\[
625a^4=(25a^2)^2 \quad \text{and} \quad 81b^8=(9b^4)^2.
\]
Thus, the expression is a difference of two squares and can be factorised as
\[
(25a^2-9b^4)(25a^2+9b^4).
\]
#### 1.2 General Statement
Any expression of the form
\[
A^2-B^2
\]
can be factorised as
\[
(A-B)(A+B).
\]
Both expressions above are examples of this rule.
---
### QUESTION 2
#### 2.1 Factorise fully and simplify where possible
**2.1.1**
We have
\[
x^2+2xy+y^2.
\]
This is a perfect square trinomial, since
\[
x^2+2xy+y^2=(x+y)^2.
\]
**2.1.2**
We have
\[
x^2-8x+16.
\]
Recognise that
\[
x^2-8x+16=(x-4)^2.
\]
**2.1.3**
We have
\[
16a^4-24a^2b^3+9b^6.
\]
Notice that
\[
(4a^2-3b^3)^2 = 16a^4 - 2\cdot(4a^2)(3b^3) + 9b^6 = 16a^4-24a^2b^3+9b^6.
\]
Thus,
\[
16a^4-24a^2b^3+9b^6=(4a^2-3b^3)^2.
\]
#### 2.2 General Statement
A perfect square trinomial of the form
\[
A^2 \pm 2AB + B^2
\]
can be factorised as
\[
(A \pm B)^2.
\]
The expressions above are examples of this complete square factorisation.
---
### QUESTION 3
#### 3.1 Factorise fully and simplify where possible
**3.1.1**
We have
\[
x^3+y^3.
\]
This is a sum of cubes, which factorises as
\[
x^3+y^3=(x+y)(x^2-xy+y^2).
\]
**3.1.2**
We have
\[
8a^6+729.
\]
Rewrite each term as a cube:
\[
8a^6 = (2a^2)^3 \quad \text{and} \quad 729 = 9^3.
\]
Using the sum of cubes formula
\[
A^3+B^3=(A+B)(A^2-AB+B^2),
\]
with \(A=2a^2\) and \(B=9\), we obtain:
\[
8a^6+729=(2a^2+9)\Big((2a^2)^2-2a^2\cdot9+9^2\Big).
\]
Simplify the second factor:
\[
(2a^2)^2=4a^4,\quad 2a^2\cdot9=18a^2,\quad 9^2=81.
\]
Thus,
\[
8a^6+729=(2a^2+9)(4a^4-18a^2+81).
\]
**3.1.3**
We have
\[
(3x+y)^3+(3x-y)^3.
\]
Let
\[
A=3x+y \quad \text{and} \quad B=3x-y.
\]
Then by the sum of cubes formula:
\[
A^3+B^3=(A+B)(A^2-AB+B^2).
\]
First, compute \(A+B\):
\[
(3x+y)+(3x-y)=6x.
\]
Now, compute \(A^2-AB+B^2\). An alternative approach is to notice that the sum of cubes identity can be expressed as
\[
A^3+B^3 = (A+B)^3-3AB(A+B).
\]
We already have \(A+B=6x\). Next, find \(AB\):
\[
AB=(3x+y)(3x-y)=9x^2-y^2.
\]
Thus,
\[
(3x+y)^3+(3x-y)^3 = (6x)^3-3(9x^2-y^2)(6x).
\]
Compute each term:
\[
(6x)^3=216x^3,
\]
\[
3(9x^2-y^2)(6x)=18x(9x^2-y^2).
\]
So,
\[
216x^3-18x(9x^2-y^2)=216x^3-162x^3+18xy^2=54x^3+18xy^2.
\]
Factor out \(18x\):
\[
54x^3+18xy^2=18x(3x^2+y^2).
\]
Thus,
\[
(3x+y)^3+(3x-y)^3 = 18x(3x^2+y^2).
\]
---
### Final Answers
**QUESTION 1:**
1.1.1 \(\quad x^2-y^2=(x-y)(x+y)\)
1.1.2 \(\quad 625a^4-81b^8=(25a^2-9b^4)(25a^2+9b^4)\)
1.2 General Statement: For any \(A^2-B^2\), factorise as \((A-B)(A+B)\).
**QUESTION 2:**
2.1.1 \(\quad x^2+2xy+y^2=(x+y)^2\)
2.1.2 \(\quad x^2-8x+16=(x-4)^2\)
2.1.3 \(\quad 16a^4-24a^2b^3+9b^6=(4a^2-3b^3)^2\)
2.2 General Statement: A perfect square trinomial \(A^2\pm2AB+B^2\) factorises as \((A\pm B)^2\).
**QUESTION 3:**
3.1.1 \(\quad x^3+y^3=(x+y)(x^2-xy+y^2)\)
3.1.2 \(\quad 8a^6+729=(2a^2+9)(4a^4-18a^2+81)\)
3.1.3 \(\quad (3x+y)^3+(3x-y)^3=18x(3x^2+y^2)\)
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