Find $ \left(a-x\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ a } $ and $ B = \color{red}{ x }$.

$$ \begin{aligned}\left(a-x\right)^2 = \color{blue}{a^2} -2 \cdot a \cdot x + \color{red}{x^2} = a^2-2ax+x^2\end{aligned} $$

Multiply $ \color{blue}{a^2x} $ by $ \left( a-x\right) $ $$ \color{blue}{a^2x} \cdot \left( a-x\right) = a^3x-a^2x^2 $$Multiply $ \color{blue}{2ax^2} $ by $ \left( a-x\right) $ $$ \color{blue}{2ax^2} \cdot \left( a-x\right) = 2a^2x^2-2ax^3 $$Multiply $ \color{blue}{ax} $ by $ \left( a^2-2ax+x^2\right) $ $$ \color{blue}{ax} \cdot \left( a^2-2ax+x^2\right) = a^3x-2a^2x^2+ax^3 $$

Multiply $a^3$ by $ \dfrac{a-x}{3} $ to get $ \dfrac{ a^4-a^3x }{ 3 } $.

Step 1: Write $ a^3 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} a^3 \cdot \frac{a-x}{3} & \xlongequal{\text{Step 1}} \frac{a^3}{\color{red}{1}} \cdot \frac{a-x}{3} \xlongequal{\text{Step 2}} \frac{ a^3 \cdot \left( a-x \right) }{ 1 \cdot 3 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ a^4-a^3x }{ 3 } \end{aligned} $$

Find $ \left(a-x\right)^2 $ using formula.

$$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$

where $ A = \color{blue}{ a } $ and $ B = \color{red}{ x }$.

$$ \begin{aligned}\left(a-x\right)^2 = \color{blue}{a^2} -2 \cdot a \cdot x + \color{red}{x^2} = a^2-2ax+x^2\end{aligned} $$

Subtract $a^3x-a^2x^2$ from $ \dfrac{a^4-a^3x}{3} $ to get $ \dfrac{ \color{purple}{ a^4-4a^3x+3a^2x^2 } }{ 3 }$.

Step 1: Write $ a^3x-a^2x^2 $ 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 second fraction by $\color{blue}{3}$.

$$ \begin{aligned} \frac{a^4-a^3x}{3} -a^3x-a^2x^2 & \xlongequal{\text{Step 1}} \frac{a^4-a^3x}{3} - \frac{a^3x-a^2x^2}{\color{red}{1}} = \frac{ a^4-a^3x }{ 3 } - \frac{ \left( a^3x-a^2x^2 \right) \cdot \color{blue}{ 3 }}{ 1 \cdot \color{blue}{ 3 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ a^4-a^3x } }{ 3 } - \frac{ \color{purple}{ 3a^3x-3a^2x^2 } }{ 3 }=\frac{ \color{purple}{ a^4-a^3x - \left( 3a^3x-3a^2x^2 \right) } }{ 3 } = \\[1ex] &=\frac{ \color{purple}{ a^4-4a^3x+3a^2x^2 } }{ 3 } \end{aligned} $$

Multiply $a^2$ by $ \dfrac{a^2-2ax+x^2}{2} $ to get $ \dfrac{ a^4-2a^3x+a^2x^2 }{ 2 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} a^2 \cdot \frac{a^2-2ax+x^2}{2} & \xlongequal{\text{Step 1}} \frac{a^2}{\color{red}{1}} \cdot \frac{a^2-2ax+x^2}{2} \xlongequal{\text{Step 2}} \frac{ a^2 \cdot \left( a^2-2ax+x^2 \right) }{ 1 \cdot 2 } = \\[1ex] & \xlongequal{\text{Step 3}} \frac{ a^4-2a^3x+a^2x^2 }{ 2 } \end{aligned} $$

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

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}{ 2 }$ and the second by $\color{blue}{ 3 }$.

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

Add $ \dfrac{-a^4-2a^3x+3a^2x^2}{6} $ and $ 2a^2x^2-2ax^3 $ to get $ \dfrac{ \color{purple}{ -a^4-2a^3x+15a^2x^2-12ax^3 } }{ 6 }$.

Step 1: Write $ 2a^2x^2-2ax^3 $ 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 $\color{blue}{6}$.

$$ \begin{aligned} \frac{-a^4-2a^3x+3a^2x^2}{6} +2a^2x^2-2ax^3 & \xlongequal{\text{Step 1}} \frac{-a^4-2a^3x+3a^2x^2}{6} + \frac{2a^2x^2-2ax^3}{\color{red}{1}} = \\[1ex] &= \frac{ -a^4-2a^3x+3a^2x^2 }{ 6 } + \frac{ \left( 2a^2x^2-2ax^3 \right) \cdot \color{blue}{ 6 }}{ 1 \cdot \color{blue}{ 6 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -a^4-2a^3x+3a^2x^2 } }{ 6 } + \frac{ \color{purple}{ 12a^2x^2-12ax^3 } }{ 6 }=\frac{ \color{purple}{ -a^4-2a^3x+15a^2x^2-12ax^3 } }{ 6 } \end{aligned} $$

Add $ \dfrac{-a^4-2a^3x+15a^2x^2-12ax^3}{6} $ and $ a^3x-2a^2x^2+ax^3 $ to get $ \dfrac{ \color{purple}{ -a^4+4a^3x+3a^2x^2-6ax^3 } }{ 6 }$.

Step 1: Write $ a^3x-2a^2x^2+ax^3 $ 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 $\color{blue}{6}$.

$$ \begin{aligned} \frac{-a^4-2a^3x+15a^2x^2-12ax^3}{6} +a^3x-2a^2x^2+ax^3 & \xlongequal{\text{Step 1}} \frac{-a^4-2a^3x+15a^2x^2-12ax^3}{6} + \frac{a^3x-2a^2x^2+ax^3}{\color{red}{1}} = \\[1ex] &= \frac{ -a^4-2a^3x+15a^2x^2-12ax^3 }{ 6 } + \frac{ \left( a^3x-2a^2x^2+ax^3 \right) \cdot \color{blue}{ 6 }}{ 1 \cdot \color{blue}{ 6 }} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ -a^4-2a^3x+15a^2x^2-12ax^3 } }{ 6 } + \frac{ \color{purple}{ 6a^3x-12a^2x^2+6ax^3 } }{ 6 } = \\[1ex] &=\frac{ \color{purple}{ -a^4+4a^3x+3a^2x^2-6ax^3 } }{ 6 } \end{aligned} $$