Find $ \left(1-a\right)^2 $ using formula.

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

where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ a }$.

$$ \begin{aligned}\left(1-a\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot a + \color{red}{a^2} = 1-2a+a^2\end{aligned} $$

Find $ \left(1-a\right)^3 $ using formula

$$ (A - B) = A^3 - 3A^2B + 3AB^2 - B^3 $$

where $ A = 1 $ and $ B = a $.

$$ \left(1-a\right)^3 = 1^3-3 \cdot 1^2 \cdot a + 3 \cdot 1 \cdot a^2-a^3 = 1-3a+3a^2-a^3 $$

$$ (1-a)^4 = (1-a)^2 \cdot (1-a)^2 $$

Find $ \left(1-a\right)^2 $ using formula.

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

where $ A = \color{blue}{ 1 } $ and $ B = \color{red}{ a }$.

$$ \begin{aligned}\left(1-a\right)^2 = \color{blue}{1^2} -2 \cdot 1 \cdot a + \color{red}{a^2} = 1-2a+a^2\end{aligned} $$

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

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

Combine like terms:

$$ 1 \color{blue}{-2a} + \color{red}{a^2} \color{blue}{-2a} + \color{green}{4a^2} \color{orange}{-2a^3} + \color{green}{a^2} \color{orange}{-2a^3} +a^4 = a^4 \color{orange}{-4a^3} + \color{green}{6a^2} \color{blue}{-4a} +1 $$

Subtract $ \dfrac{1-2a+a^2}{2a^2} $ from $ \dfrac{1-a}{a} $ to get $ \dfrac{ \color{purple}{ -3a^3+4a^2-a } }{ 2a^3 }$.

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

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

Multiply $2$ by $ \dfrac{1-3a+3a^2-a^3}{6a^3} $ to get $ \dfrac{-2a^3+6a^2-6a+2}{6a^3} $.

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.

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

Multiply $6$ by $ \dfrac{a^4-4a^3+6a^2-4a+1}{24a^4} $ to get $ \dfrac{ 6a^4-24a^3+36a^2-24a+6 }{ 24a^4 } $.

Step 1: Write $ 6 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 6 \cdot \frac{a^4-4a^3+6a^2-4a+1}{24a^4} & \xlongequal{\text{Step 1}} \frac{6}{\color{red}{1}} \cdot \frac{a^4-4a^3+6a^2-4a+1}{24a^4} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 6 \cdot \left( a^4-4a^3+6a^2-4a+1 \right) }{ 1 \cdot 24a^4 } \xlongequal{\text{Step 3}} \frac{ 6a^4-24a^3+36a^2-24a+6 }{ 24a^4 } \end{aligned} $$

Add $ \dfrac{-3a^3+4a^2-a}{2a^3} $ and $ \dfrac{-2a^3+6a^2-6a+2}{6a^3} $ to get $ \dfrac{ \color{purple}{ -22a^6+36a^5-18a^4+4a^3 } }{ 12a^6 }$.

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}{ 6a^3 }$ and the second by $\color{blue}{ 2a^3 }$.

$$ \begin{aligned} \frac{-3a^3+4a^2-a}{2a^3} + \frac{-2a^3+6a^2-6a+2}{6a^3} & = \frac{ \left( -3a^3+4a^2-a \right) \cdot \color{blue}{ 6a^3 }}{ 2a^3 \cdot \color{blue}{ 6a^3 }} + \frac{ \left( -2a^3+6a^2-6a+2 \right) \cdot \color{blue}{ 2a^3 }}{ 6a^3 \cdot \color{blue}{ 2a^3 }} = \\[1ex] &=\frac{ \color{purple}{ -18a^6+24a^5-6a^4 } }{ 12a^6 } + \frac{ \color{purple}{ -4a^6+12a^5-12a^4+4a^3 } }{ 12a^6 } = \\[1ex] &=\frac{ \color{purple}{ -22a^6+36a^5-18a^4+4a^3 } }{ 12a^6 } \end{aligned} $$

Multiply $6$ by $ \dfrac{a^4-4a^3+6a^2-4a+1}{24a^4} $ to get $ \dfrac{ 6a^4-24a^3+36a^2-24a+6 }{ 24a^4 } $.

Step 1: Write $ 6 $ as a fraction by putting $ \color{red}{1} $ in the denominator.

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

$$ \begin{aligned} 6 \cdot \frac{a^4-4a^3+6a^2-4a+1}{24a^4} & \xlongequal{\text{Step 1}} \frac{6}{\color{red}{1}} \cdot \frac{a^4-4a^3+6a^2-4a+1}{24a^4} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ 6 \cdot \left( a^4-4a^3+6a^2-4a+1 \right) }{ 1 \cdot 24a^4 } \xlongequal{\text{Step 3}} \frac{ 6a^4-24a^3+36a^2-24a+6 }{ 24a^4 } \end{aligned} $$

Subtract $ \dfrac{6a^4-24a^3+36a^2-24a+6}{24a^4} $ from $ \dfrac{-22a^6+36a^5-18a^4+4a^3}{12a^6} $ to get $ \dfrac{ \color{purple}{ -600a^{10}+1152a^9-864a^8+384a^7-72a^6 } }{ 288a^{10} }$.

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}{ 24a^4 }$ and the second by $\color{blue}{ 12a^6 }$.

$$ \begin{aligned} \frac{-22a^6+36a^5-18a^4+4a^3}{12a^6} - \frac{6a^4-24a^3+36a^2-24a+6}{24a^4} & = \frac{ \left( -22a^6+36a^5-18a^4+4a^3 \right) \cdot \color{blue}{ 24a^4 }}{ 12a^6 \cdot \color{blue}{ 24a^4 }} - \frac{ \left( 6a^4-24a^3+36a^2-24a+6 \right) \cdot \color{blue}{ 12a^6 }}{ 24a^4 \cdot \color{blue}{ 12a^6 }} = \\[1ex] &=\frac{ \color{purple}{ -528a^{10}+864a^9-432a^8+96a^7 } }{ 288a^{10} } - \frac{ \color{purple}{ 72a^{10}-288a^9+432a^8-288a^7+72a^6 } }{ 288a^{10} } = \\[1ex] &=\frac{ \color{purple}{ -528a^{10}+864a^9-432a^8+96a^7 - \left( 72a^{10}-288a^9+432a^8-288a^7+72a^6 \right) } }{ 288a^{10} } = \\[1ex] &=\frac{ \color{purple}{ -600a^{10}+1152a^9-864a^8+384a^7-72a^6 } }{ 288a^{10} } \end{aligned} $$