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Question

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

How do you simplify (y4)3{{\left( y-4 \right)}^{3}}?

Explanation

Solution

We first find the simplification of the given polynomial (y4)3{{\left( y-4 \right)}^{3}} according to the identity (ab)3=a33a2b+3ab2b3{{\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 (ab)2=a2+b22ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab. We then multiply the term (ab)\left( a-b \right) on both sides of the identity. We solve the multiplication to find the simplified form of (y4)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 (y4)3{{\left( y-4 \right)}^{3}}. This is the cube of difference of two numbers. We know that (ab)2=a2+b22ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
We need to multiply the term (ab)\left( a-b \right) on both side of the identity (ab)2=a2+b22ab{{\left( a-b \right)}^{2}}={{a}^{2}}+{{b}^{2}}-2ab.
On the left side of the equation, we get (ab)2(ab)=(ab)3{{\left( a-b \right)}^{2}}\left( a-b \right)={{\left( a-b \right)}^{3}}.
On the right side we have (a2+b22ab)(ab)\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\left( a-b \right). We use multiplication and get
(a2+b22ab)(ab) =a2.a+a.b22ab×aa2.bb2.b+2ab.b =a3+ab22a2ba2bb3+2ab2 =a33a2b+3ab2b3 \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
(ab)3=a33a2b+3ab2b3=a3b33ab(ab){{\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 (ab)3=a33a2b+3ab2b3{{\left( a-b \right)}^{3}}={{a}^{3}}-3{{a}^{2}}b+3a{{b}^{2}}-{{b}^{3}}.
(y4)3 =y33y2×4+3y×4243 =y312y2+48y64 \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 (y4)3{{\left( y-4 \right)}^{3}} is y312y2+48y64{{y}^{3}}-12{{y}^{2}}+48y-64.
Now we verify the result with an arbitrary value of y=2y=2.
We have (y4)3=y312y2+48y64{{\left( y-4 \right)}^{3}}={{y}^{3}}-12{{y}^{2}}+48y-64.
The left-hand side of the equation gives (y4)3=(24)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
y312y2+48y64 =2312×22+48×264 =848+9664 =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 (y4)3=y312y2+48y64{{\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 (ab)n=nC0anb0nC1an1b1+nC2an2b2....+(1)rnCranrbr+....+(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=3C0a3b03C1a31b1+3C2a32b23C3a33b3=a33a2b+3ab2b3{{\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}}.