Divide $ \dfrac{-7}{3} $ by $ x $ to get $ \dfrac{ -7 }{ 3x } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{-7}{3} }{x} & \xlongequal{\text{Step 1}} \frac{-7}{3} \cdot \frac{\color{blue}{1}}{\color{blue}{x}} \xlongequal{\text{Step 2}} \frac{ \left( -7 \right) \cdot 1 }{ 3 \cdot x } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ -7 }{ 3x } \end{aligned} $$

Multiply $ \dfrac{7}{3} $ by $ x $ to get $ \dfrac{ 7x }{ 3 } $.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{7}{3} \cdot x & \xlongequal{\text{Step 1}} \frac{7}{3} \cdot \frac{x}{\color{red}{1}} \xlongequal{\text{Step 2}} \frac{ 7 \cdot x }{ 3 \cdot 1 } \xlongequal{\text{Step 3}} \frac{ 7x }{ 3 } \end{aligned} $$

Multiply each term of $ \left( \color{blue}{x^2+x+1}\right) $ by each term in $ \left( x^2+x+1\right) $.

$$ \left( \color{blue}{x^2+x+1}\right) \cdot \left( x^2+x+1\right) = x^4+x^3+x^2+x^3+x^2+x+x^2+x+1 $$

Combine like terms:

$$ x^4+ \color{blue}{x^3} + \color{red}{x^2} + \color{blue}{x^3} + \color{green}{x^2} + \color{orange}{x} + \color{green}{x^2} + \color{orange}{x} +1 = x^4+ \color{blue}{2x^3} + \color{green}{3x^2} + \color{orange}{2x} +1 $$

Add $ \dfrac{-7}{3x} $ and $ \dfrac{1}{x^2} $ to get $ \dfrac{ \color{purple}{ -7x^2+3x } }{ 3x^3 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x^2 }$ and the second by $\color{blue}{ 3x }$.

$$ \begin{aligned} \frac{-7}{3x} + \frac{1}{x^2} & = \frac{ \left( -7 \right) \cdot \color{blue}{ x^2 }}{ 3x \cdot \color{blue}{ x^2 }} + \frac{ 1 \cdot \color{blue}{ 3x }}{ x^2 \cdot \color{blue}{ 3x }} = \\[1ex] &=\frac{ \color{purple}{ -7x^2 } }{ 3x^3 } + \frac{ \color{purple}{ 3x } }{ 3x^3 }=\frac{ \color{purple}{ -7x^2+3x } }{ 3x^3 } \end{aligned} $$

Add $ \dfrac{7x}{3} $ and $ 1 $ to get $ \dfrac{ \color{purple}{ 7x+3 } }{ 3 }$.

Step 1: Write $ 1 $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{7x}{3} +1 & \xlongequal{\text{Step 1}} \frac{7x}{3} + \frac{1}{\color{red}{1}} = \frac{ 7x }{ 3 } + \frac{ 1 \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 7x } }{ 3 } + \frac{ \color{purple}{ 3 } }{ 3 }=\frac{ \color{purple}{ 7x+3 } }{ 3 } \end{aligned} $$

Multiply each term of $ \left( \color{blue}{x^2+x+1}\right) $ by each term in $ \left( x^2+x+1\right) $.

$$ \left( \color{blue}{x^2+x+1}\right) \cdot \left( x^2+x+1\right) = x^4+x^3+x^2+x^3+x^2+x+x^2+x+1 $$

Combine like terms:

$$ x^4+ \color{blue}{x^3} + \color{red}{x^2} + \color{blue}{x^3} + \color{green}{x^2} + \color{orange}{x} + \color{green}{x^2} + \color{orange}{x} +1 = x^4+ \color{blue}{2x^3} + \color{green}{3x^2} + \color{orange}{2x} +1 $$

Add $ \dfrac{-7}{3x} $ and $ \dfrac{1}{x^2} $ to get $ \dfrac{ \color{purple}{ -7x^2+3x } }{ 3x^3 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x^2 }$ and the second by $\color{blue}{ 3x }$.

$$ \begin{aligned} \frac{-7}{3x} + \frac{1}{x^2} & = \frac{ \left( -7 \right) \cdot \color{blue}{ x^2 }}{ 3x \cdot \color{blue}{ x^2 }} + \frac{ 1 \cdot \color{blue}{ 3x }}{ x^2 \cdot \color{blue}{ 3x }} = \\[1ex] &=\frac{ \color{purple}{ -7x^2 } }{ 3x^3 } + \frac{ \color{purple}{ 3x } }{ 3x^3 }=\frac{ \color{purple}{ -7x^2+3x } }{ 3x^3 } \end{aligned} $$

Divide $ \dfrac{7x+3}{3} $ by $ x^2+x+1 $ to get $ \dfrac{ 7x+3 }{ 3x^2+3x+3 } $.

Step 1: To divide rational expressions, multiply the first fraction by the reciprocal of the second fraction.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} \frac{ \frac{7x+3}{3} }{x^2+x+1} & \xlongequal{\text{Step 1}} \frac{7x+3}{3} \cdot \frac{\color{blue}{1}}{\color{blue}{x^2+x+1}} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 7x+3 \right) \cdot 1 }{ 3 \cdot \left( x^2+x+1 \right) } \xlongequal{\text{Step 3}} \frac{ 7x+3 }{ 3x^2+3x+3 } \end{aligned} $$

Multiply each term of $ \left( \color{blue}{x^2+x+1}\right) $ by each term in $ \left( x^2+x+1\right) $.

$$ \left( \color{blue}{x^2+x+1}\right) \cdot \left( x^2+x+1\right) = x^4+x^3+x^2+x^3+x^2+x+x^2+x+1 $$

Combine like terms:

$$ x^4+ \color{blue}{x^3} + \color{red}{x^2} + \color{blue}{x^3} + \color{green}{x^2} + \color{orange}{x} + \color{green}{x^2} + \color{orange}{x} +1 = x^4+ \color{blue}{2x^3} + \color{green}{3x^2} + \color{orange}{2x} +1 $$

Add $ \dfrac{-7x^2+3x}{3x^3} $ and $ \dfrac{7x+3}{3x^2+3x+3} $ to get $ \dfrac{ \color{purple}{ -3x^3-12x^2+9x } }{ 9x^5+9x^4+9x^3 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ 3x^2+3x+3 }$ and the second by $\color{blue}{ 3x^3 }$.

$$ \begin{aligned} \frac{-7x^2+3x}{3x^3} + \frac{7x+3}{3x^2+3x+3} & = \frac{ \left( -7x^2+3x \right) \cdot \color{blue}{ \left( 3x^2+3x+3 \right) }}{ 3x^3 \cdot \color{blue}{ \left( 3x^2+3x+3 \right) }} + \frac{ \left( 7x+3 \right) \cdot \color{blue}{ 3x^3 }}{ \left( 3x^2+3x+3 \right) \cdot \color{blue}{ 3x^3 }} = \\[1ex] &=\frac{ \color{purple}{ -21x^4-21x^3-21x^2+9x^3+9x^2+9x } }{ 9x^5+9x^4+9x^3 } + \frac{ \color{purple}{ 21x^4+9x^3 } }{ 9x^5+9x^4+9x^3 } = \\[1ex] &=\frac{ \color{purple}{ -3x^3-12x^2+9x } }{ 9x^5+9x^4+9x^3 } \end{aligned} $$

Multiply each term of $ \left( \color{blue}{x^2+x+1}\right) $ by each term in $ \left( x^2+x+1\right) $.

$$ \left( \color{blue}{x^2+x+1}\right) \cdot \left( x^2+x+1\right) = x^4+x^3+x^2+x^3+x^2+x+x^2+x+1 $$

Combine like terms:

$$ x^4+ \color{blue}{x^3} + \color{red}{x^2} + \color{blue}{x^3} + \color{green}{x^2} + \color{orange}{x} + \color{green}{x^2} + \color{orange}{x} +1 = x^4+ \color{blue}{2x^3} + \color{green}{3x^2} + \color{orange}{2x} +1 $$

Subtract $ \dfrac{x+1}{x^4+2x^3+3x^2+2x+1} $ from $ \dfrac{-x^2-4x+3}{3x^4+3x^3+3x^2} $ to get $ \dfrac{ \color{purple}{ -x^4-8x^3-5x^2-x+3 } }{ 3x^6+6x^5+9x^4+6x^3+3x^2 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $ \color{blue}{ x^2+x+1 }$ and the second by $\color{blue}{ 3x^2 }$.

$$ \begin{aligned} \frac{-x^2-4x+3}{3x^4+3x^3+3x^2} - \frac{x+1}{x^4+2x^3+3x^2+2x+1} & = \frac{ \left( -x^2-4x+3 \right) \cdot \color{blue}{ \left( x^2+x+1 \right) }}{ \left( 3x^4+3x^3+3x^2 \right) \cdot \color{blue}{ \left( x^2+x+1 \right) }} - \frac{ \left( x+1 \right) \cdot \color{blue}{ 3x^2 }}{ \left( x^4+2x^3+3x^2+2x+1 \right) \cdot \color{blue}{ 3x^2 }} = \\[1ex] &=\frac{ \color{purple}{ -x^4-x^3-x^2-4x^3-4x^2-4x+3x^2+3x+3 } }{ 3x^6+3x^5+3x^4+3x^5+3x^4+3x^3+3x^4+3x^3+3x^2 } - \frac{ \color{purple}{ 3x^3+3x^2 } }{ 3x^6+3x^5+3x^4+3x^5+3x^4+3x^3+3x^4+3x^3+3x^2 } = \\[1ex] &=\frac{ \color{purple}{ -x^4-x^3-x^2-4x^3-4x^2-4x+3x^2+3x+3 - \left( 3x^3+3x^2 \right) } }{ 3x^6+6x^5+9x^4+6x^3+3x^2 } = \\[1ex] &=\frac{ \color{purple}{ -x^4-8x^3-5x^2-x+3 } }{ 3x^6+6x^5+9x^4+6x^3+3x^2 } \end{aligned} $$