ca. $$ (x+3)\left(x^{2}-3 x+9\right) $$ (b) $$ \left(4 a^{2}-2 a b+b^{2}\right)(2 a+b) $$ c) $$ (a-2)\left(a^{2}+2 a+6\right) $$ d, $$ \left(\frac{9}{2}+1\right)\left(\frac{9}{2}-1\right) $$ i. $$ (3 a+5)^{2} $$ if, $$ (2 x-5)^{2} $$ (s) $$ 3 x(2 x-4 x y) $$ h) $$ a b^{2}\left(-2 a^{2}+4 b\right) $$ I $$ \left(x-\frac{1}{x}\right)^{2} $$ i. $$ \left(x-\frac{1}{x}\right)\left(x^{2}+1+\frac{1}{x^{2}}\right) $$

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UpStudy AI · Accepted Answer

Sure! Let's go through each of the expressions you've provided and simplify or factor them as needed.

---

### **a. $$ (x + 3)\left(x^{2} - 3x + 9\right) $$**

This expression is already factored. However, we can recognize that $$x^{3} + 27$$ can be factored using the sum of cubes formula:

$$
x^{3} + 27 = (x + 3)(x^{2} - 3x + 9)
$$

So,

$$
(x + 3)\left(x^{2} - 3x + 9\right) = x^{3} + 27
$$

---

### **b. $$ \left(4a^{2} - 2ab + b^{2}\right)(2a + b) $$**

First, let's factor the quadratic $$4a^{2} - 2ab + b^{2}$$:

This quadratic can be written as:

$$
4a^{2} - 2ab + b^{2} = (2a)^2 - 2a \cdot b + b^2 = (2a - b)^2
$$

So,

$$
\left(4a^{2} - 2ab + b^{2}\right)(2a + b) = (2a - b)^2 (2a + b)
$$

This is the fully factored form.

---

### **c. $$ (a - 2)\left(a^{2} + 2a + 6\right) $$**

Let's attempt to factor $$a^{2} + 2a + 6$$:

Calculate the discriminant:

$$
\Delta = (2)^2 - 4 \cdot 1 \cdot 6 = 4 - 24 = -20
$$

Since the discriminant is negative, the quadratic does not factor over the real numbers. Therefore, the expression remains:

$$
(a - 2)\left(a^{2} + 2a + 6\right)
$$

If complex factors are acceptable, it can be written as:

$$
(a - 2)\left(a + 1 + i\sqrt{5}\right)\left(a + 1 - i\sqrt{5}\right)
$$

---

### **d. $$ \left(\frac{9}{2} + 1\right)\left(\frac{9}{2} - 1\right) $$**

Simplify each term:

$$
\frac{9}{2} + 1 = \frac{9}{2} + \frac{2}{2} = \frac{11}{2}
$$
$$
\frac{9}{2} - 1 = \frac{9}{2} - \frac{2}{2} = \frac{7}{2}
$$

Multiply them:

$$
\frac{11}{2} \times \frac{7}{2} = \frac{77}{4}
$$

---

### **e. $$ (3a + 5)^{2} $$**

Expand the square:

$$
(3a + 5)^2 = (3a)^2 + 2 \cdot 3a \cdot 5 + 5^2 = 9a^2 + 30a + 25
$$

Alternatively, if further factoring is possible, but in this case, it remains as is.

---

### **f. $$ (2x - 5)^{2} $$**

Expand the square:

$$
(2x - 5)^2 = (2x)^2 - 2 \cdot 2x \cdot 5 + 5^2 = 4x^2 - 20x + 25
$$

---

### **g. $$ 3x(2x - 4xy) $$**

Factor out the common terms inside the parentheses:

$$
2x - 4xy = 2x(1 - 2y)
$$

So,

$$
3x(2x - 4xy) = 3x \cdot 2x (1 - 2y) = 6x^2(1 - 2y)
$$

---

### **h. $$ a b^{2}\left(-2a^{2} + 4b\right) $$**

Factor out the common factor inside the parentheses:

$$
-2a^{2} + 4b = -2(a^{2} - 2b)
$$

So,

$$
a b^{2}\left(-2a^{2} + 4b\right) = -2a b^{2}(a^{2} - 2b)
$$

---

### **i. $$ \left(x - \frac{1}{x}\right)^{2} $$**

Expand the square:

$$
\left(x - \frac{1}{x}\right)^2 = x^2 - 2 \cdot x \cdot \frac{1}{x} + \left(\frac{1}{x}\right)^2 = x^2 - 2 + \frac{1}{x^2}
$$

---

### **j. $$ \left(x - \frac{1}{x}\right)\left(x^{2} + 1 + \frac{1}{x^{2}}\right) $$**

Multiply the two expressions:

$$
\left(x - \frac{1}{x}\right)\left(x^{2} + 1 + \frac{1}{x^{2}}\right)
$$

Multiply term by term:

$$
x \cdot x^2 = x^3
$$
$$
x \cdot 1 = x
$$
$$
x \cdot \frac{1}{x^2} = \frac{1}{x}
$$
$$
-\frac{1}{x} \cdot x^2 = -x
$$
$$
-\frac{1}{x} \cdot 1 = -\frac{1}{x}
$$
$$
-\frac{1}{x} \cdot \frac{1}{x^2} = -\frac{1}{x^3}
$$

Combine all terms:

$$
x^3 + x + \frac{1}{x} - x - \frac{1}{x} - \frac{1}{x^3} = x^3 - \frac{1}{x^3}
$$

So,

$$
\left(x - \frac{1}{x}\right)\left(x^{2} + 1 + \frac{1}{x^{2}}\right) = x^3 - \frac{1}{x^3}
$$

---

Feel free to ask if you need further clarification or assistance with any of these problems!
Here are the simplified forms of the given expressions: a. $$ (x + 3)\left(x^{2} - 3x + 9\right) = x^{3} + 27 $$ b. $$ \left(4a^{2} - 2ab + b^{2}\right)(2a + b) = (2a - b)^2 (2a + b) $$ c. $$ (a - 2)\left(a^{2} + 2a + 6\right) $$ (Cannot be factored further over real numbers) d. $$ \left(\frac{9}{2} + 1\right)\left(\frac{9}{2} - 1\right) = \frac{77}{4} $$ e. $$ (3a + 5)^{2} = 9a^{2} + 30a + 25 $$ f. $$ (2x - 5)^{2} = 4x^{2} - 20x + 25 $$ g. $$ 3x(2x - 4xy) = 6x^{2}(1 - 2y) $$ h. $$ a b^{2}\left(-2a^{2} + 4b\right) = -2a b^{2}(a^{2} - 2b) $$ i. $$ \left(x - \frac{1}{x}\right)^{2} = x^{2} - 2 + \frac{1}{x^{2}} $$ j. $$ \left(x - \frac{1}{x}\right)\left(x^{2} + 1 + \frac{1}{x^{2}}\right) = x^{3} - \frac{1}{x^{3}} $$

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