Answer

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**Hint:**We need to solve this question by using binomial theorem so that we can find each term here. In this theorem, we expand the polynomial \[{(x + y)^n}\] in a sum which is having the terms in form of \[a{x^b}{y^c}\] where \[b\,and\,c\] are with integers that are not negative with \[b + c = n\] and \[a\] of every term is positive depending on \[n\,and\,b\].

**Formula used:**

\[{\left( {a + b} \right)^n} = \mathop \sum \limits_{k = 0}^n nCk \cdot \left( {{a^{n - k}}{b^k}} \right)\]

**Complete step by step solution:**

By using binomial theorem, we are going to find each term. The formula for binomial theorem is:

\[{\left( {a + b} \right)^n} = \mathop \sum \limits_{k = 0}^n nCk \cdot \left( {{a^{n - k}}{b^k}} \right)\]

According to this formula we can get that \[n\, = \,3\], \[a\, = \,x\], \[b\, = - 2\]. So, after getting the values according to the questions, apply the values in the formula and after that we get:

\[\sum\limits_{k = 0}^3 {\dfrac{{3!}}{{(3 - k)!k!}} \cdot {{(x)}^{3 - k}} \cdot {{( - 2)}^k}} \]

Now, we will start expanding the whole summation, and we will put the value of \[k\] as \[0,1,2,3\] and then we get:

\[ \Rightarrow \dfrac{{3!}}{{(3 - 0)!0!}} \cdot {(x)^{3 - 0}} \cdot {( - 2)^0} + \dfrac{{3!}}{{(3 - 1)!1!}} \cdot {(x)^{3 - 1}} \cdot {( - 2)^1} + \dfrac{{3!}}{{(3 - 2)!2!}} \cdot {(x)^{3 - 2}} \cdot {( - 2)^2} + \dfrac{{3!}}{{(3 - 3)!3!}} \cdot {(x)^{3 - 3}} \cdot {( - 2)^3}\]

Now, by simplifying all the exponents for every term we will get:

\[ \Rightarrow 1{(x)^3}{( - 2)^0} + 3{(x)^2}( - 2) + 3(x){( - 2)^2} + 1{(x)^0}{( - 2)^3}\]

Now, we need to simplify every term:

\[ \Rightarrow {(x)^3} \cdot {( - 2)^0} + 3 \cdot {(x)^2} \cdot {( - 2)^1} + 3 \cdot {(x)^1} \cdot {( - 2)^2} + 1 \cdot {(x)^0} \cdot {( - 2)^3}\]

\[ \therefore{x^3} - 6{(x)^2} - 12x - 8\]

**So, the final answer is \[{x^3} - 6{(x)^2} - 12x - 8\].**

**Additional information:**

Binomial number is a polynomial having two terms. Pascal’s triangle could be a triangular cluster built by summing adjoining components in going before columns. This contains the values of all the binomial coefficients.

**Note:**This method is very easy, and we can quickly get the answer by solving all the equations, but there is another method through which we can solve it and it is called Pascal’s Triangle. Through Pascal’s Triangle method, we can solve the question very easily but it might be a bit complicated.

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