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Answer

$$(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+b(y-\frac{a}{3})+c=(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+\frac{-ab+3by}{3}+c$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+b(y-\frac{a}{3})+c& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+b\frac{-a+3y}{3}+c \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(y-\frac{a}{3})^3+a(y-\frac{a}{3})^2+\frac{-ab+3by}{3}+c\end{aligned} $$
①

Subtract $ \dfrac{a}{3} $ from $ y $ to get $ \dfrac{ \color{purple}{ -a+3y } }{ 3 }$.

Step 1: Write $ y $ 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 $\color{blue}{ 3 }$.

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

Multiply $b$ by $ \dfrac{-a+3y}{3} $ to get $ \dfrac{ -ab+3by }{ 3 } $.

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

Step 2: Multiply numerators and denominators.

Step 3: Simplify numerator and denominator.

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