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# How do you expand the binomial ${(x - 2)^3}$?

Last updated date: 02nd Aug 2024
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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 ${x^3} - 6{(x)^2} - 12x - 8$.