Subtract $ \dfrac{x^3}{3} $ from $ x $ to get $ \dfrac{ \color{purple}{ -x^3+3x } }{ 3 }$.

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

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

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

Multiply $2$ by $ \dfrac{x^5}{15} $ to get $ \dfrac{ 2x^5 }{ 15 } $.

Step 1: Write $ 2 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 2 \cdot \frac{x^5}{15} & \xlongequal{\text{Step 1}} \frac{2}{\color{red}{1}} \cdot \frac{x^5}{15} \xlongequal{\text{Step 2}} \frac{ 2 \cdot x^5 }{ 1 \cdot 15 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 2x^5 }{ 15 } \end{aligned} $$

Multiply $17$ by $ \dfrac{x^7}{315} $ to get $ \dfrac{ 17x^7 }{ 315 } $.

Step 1: Write $ 17 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 17 \cdot \frac{x^7}{315} & \xlongequal{\text{Step 1}} \frac{17}{\color{red}{1}} \cdot \frac{x^7}{315} \xlongequal{\text{Step 2}} \frac{ 17 \cdot x^7 }{ 1 \cdot 315 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 17x^7 }{ 315 } \end{aligned} $$

Add $ \dfrac{-x^3+3x}{3} $ and $ \dfrac{2x^5}{15} $ to get $ \dfrac{ \color{purple}{ 2x^5-5x^3+15x } }{ 15 }$.

To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 5 }$.

$$ \begin{aligned} \frac{-x^3+3x}{3} + \frac{2x^5}{15} & = \frac{ \left( -x^3+3x \right) \cdot \color{blue}{ 5 }}{ 3 \cdot \color{blue}{ 5 }} + \frac{ 2x^5 }{ 15 } = \\[1ex] &=\frac{ \color{purple}{ -5x^3+15x } }{ 15 } + \frac{ \color{purple}{ 2x^5 } }{ 15 }=\frac{ \color{purple}{ 2x^5-5x^3+15x } }{ 15 } \end{aligned} $$

Multiply $17$ by $ \dfrac{x^7}{315} $ to get $ \dfrac{ 17x^7 }{ 315 } $.

Step 1: Write $ 17 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 17 \cdot \frac{x^7}{315} & \xlongequal{\text{Step 1}} \frac{17}{\color{red}{1}} \cdot \frac{x^7}{315} \xlongequal{\text{Step 2}} \frac{ 17 \cdot x^7 }{ 1 \cdot 315 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ 17x^7 }{ 315 } \end{aligned} $$

Subtract $ \dfrac{17x^7}{315} $ from $ \dfrac{2x^5-5x^3+15x}{15} $ to get $ \dfrac{ \color{purple}{ -17x^7+42x^5-105x^3+315x } }{ 315 }$.

To subtract raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 21 }$.

$$ \begin{aligned} \frac{2x^5-5x^3+15x}{15} - \frac{17x^7}{315} & = \frac{ \left( 2x^5-5x^3+15x \right) \cdot \color{blue}{ 21 }}{ 15 \cdot \color{blue}{ 21 }} - \frac{ 17x^7 }{ 315 } = \\[1ex] &=\frac{ \color{purple}{ 42x^5-105x^3+315x } }{ 315 } - \frac{ \color{purple}{ 17x^7 } }{ 315 }=\frac{ \color{purple}{ -17x^7+42x^5-105x^3+315x } }{ 315 } \end{aligned} $$