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

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

Explanation

Solution

We first find the simplification of the given polynomial (y−4)3{{\left( y-4 \right)}^{3}} according to the identity (a−b)3=a3−3a2b+3ab2−b3{{\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 (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab. We then multiply the term (a−b)\left( a-b \right) on both sides of the identity. We solve the multiplication to find the simplified form of (y−4)3{{\left( y-4 \right)}^{3}} by replacing with a=y;b=4a=y;b=4.

Complete answer:
We need to find the simplified form of (y−4)3{{\left( y-4 \right)}^{3}}. This is the cube of difference of two numbers. We know that (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
We need to multiply the term (a−b)\left( a-b \right) on both side of the identity (a−b)2=a2+b2−2ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
On the left side of the equation, we get (a−b)2(a−b)=(a−b)3{{\left( a-b \right)}^{2}}\left( a-b \right)={{\left( a-b \right)}^{3}}.
On the right side we have (a2+b2−2ab)(a−b)\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right). We use multiplication and get
⇒(a2+b2−2ab)(a−b) =a2.a+a.b2−2ab×a−a2.b−b2.b+2ab.b =a3+ab2−2a2b−a2b−b3+2ab2 =a3−3a2b+3ab2−b3 \begin{aligned} & \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{aligned}
We also can take another form where
(a−b)3=a3−3a2b+3ab2−b3=a3−b3−3ab(a−b){{\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=4a=y;b=4 in the equation (a−b)3=a3−3a2b+3ab2−b3{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}.
(y−4)3 =y3−3y2×4+3y×42−43 =y3−12y2+48y−64 \begin{aligned} & {{\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{aligned}
Therefore, the simplified form of (y−4)3{{\left( y-4 \right)}^{3}} is y3−12y2+48y−64{{y}^{3}}-12{{y}^{2}}+48y-64.
Now we verify the result with an arbitrary value of y=2y=2.
We have (y−4)3=y3−12y2+48y−64{{\left( y-4 \right)}^{3}}={{y}^{3}}-12{{y}^{2}}+48y-64.
The left-hand side of the equation gives (y−4)3=(2−4)3=(−2)3=−8{{\left( y-4 \right)}^{3}}={{\left( 2-4 \right)}^{3}}={{\left( -2 \right)}^{3}}=-8.
The left-hand side of the equation gives
y3−12y2+48y−64 =23−12×22+48×2−64 =8−48+96−64 =−8 \begin{aligned} & {{y}^{3}}-12{{y}^{2}}+48y-64 \\\ & ={{2}^{3}}-12\times {{2}^{2}}+48\times 2-64 \\\ & =8-48+96-64 \\\ & =-8 \\\ \end{aligned}
Thus, verified the result of (y−4)3=y3−12y2+48y−64{{\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 (a−b)n=nC0anb0−nC1an−1b1+nC2an−2b2−....+(−1)rnCran−rbr+....+(−1)nnCna0bn{{\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=3n=3.
(a+b)3=3C0a3b0−3C1a3−1b1+3C2a3−2b2−3C3a3−3b3=a3−3a2b+3ab2−b3{{\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 (a+b)3{{\left( a+b \right)}^{3}}.