Question

How do you simplify ${{\left( y-4 \right)}^{3}}$?

Answer 1

Hint: We first find the simplification of the given polynomial ${{\left( y-4 \right)}^{3}}$ according to the identity ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$. We need to simplify the cubic polynomial of the sum of two numbers. We already have the identity of ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$. We then multiply the term $\left( a-b \right)$ on both sides of the identity. We solve the multiplication to find the simplified form of ${{\left( y-4 \right)}^{3}}$ by replacing with $a=y;b=4$.

Complete answer:
We need to find the simplified form of ${{\left( y-4 \right)}^{3}}$. This is the cube of difference of two numbers. We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
We need to multiply the term $\left( a-b \right)$ on both side of the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab$.
On the left side of the equation, we get ${{\left( a-b \right)}^{2}}\left( a-b \right)={{\left( a-b \right)}^{3}}$.
On the right side we have $\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right)$. We use multiplication and get
$\begin{align}
  & \Rightarrow \left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right) \\
 & ={{a}^{2}}.a+a.{{b}^{2}}-2ab\times a-{{a}^{2}}.b-{{b}^{2}}.b+2ab.b \\
 & ={{a}^{3}}+a{{b}^{2}}-2{{a}^{2}}b-{{a}^{2}}b-{{b}^{3}}+2a{{b}^{2}} \\
 & ={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}} \\
\end{align}$
We also can take another form where
${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)$.
Now we replace the values for $a=y;b=4$ in the equation ${{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$.
$\begin{align}
  & {{\left( y-4 \right)}^{3}} \\
 & ={{y}^{3}}-3{{y}^{2}}\times 4+3y\times {{4}^{2}}-{{4}^{3}} \\
 & ={{y}^{3}}-12{{y}^{2}}+48y-64 \\
\end{align}$
Therefore, the simplified form of ${{\left( y-4 \right)}^{3}}$ is ${{y}^{3}}-12{{y}^{2}}+48y-64$.
Now we verify the result with an arbitrary value of $y=2$.
We have ${{\left( y-4 \right)}^{3}}={{y}^{3}}-12{{y}^{2}}+48y-64$.
The left-hand side of the equation gives ${{\left( y-4 \right)}^{3}}={{\left( 2-4 \right)}^{3}}={{\left( -2 \right)}^{3}}=-8$.
The left-hand side of the equation gives
$\begin{align}
  & {{y}^{3}}-12{{y}^{2}}+48y-64 \\
 & ={{2}^{3}}-12\times {{2}^{2}}+48\times 2-64 \\
 & =8-48+96-64 \\
 & =-8 \\
\end{align}$
Thus, verified the result of ${{\left( y-4 \right)}^{3}}={{y}^{3}}-12{{y}^{2}}+48y-64$.

Note: We also can use the binomial theorem to find the general form and then put the value of 3.
We have ${{\left( a-b \right)}^{n}}={}^{n}{{C}_{0}}{{a}^{n}}{{b}^{0}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{b}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{b}^{2}}-....+{{\left( -1 \right)}^{r}}{}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}+....+{{\left( -1 \right)}^{n}}{}^{n}{{C}_{n}}{{a}^{0}}{{b}^{n}}$. We need to find the cube of sum of two numbers. So, we put $n=3$.
${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}-{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}-{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}$.
In this way we also simplify the term of ${{\left( a+b \right)}^{3}}$.