Question

Expand${\left(x-2\right)}^6$using the Binomial Theorem

Short answer

The value of${\left(x-2\right)}^6=x^6-12x^5+60x^4-160x^3+240x^2-192x+64$

Step-by-step solution

Step 1. Defination

Let $x$and $a$be real numbers for any positiveinteger$n$, we have${\left(x+a\right)}^n=\left(\begin{array}{c}n\\ 0\end{array}\right)x^n+\left(\begin{array}{c}n\\ 1\end{array}\right)a{x}^{n-1}+\cdots \cdots +\left(\begin{array}{c}n\\ j\end{array}\right)a^j{x}^{n-j}+\cdots \cdots +\left(\begin{array}{c}n\\ n\end{array}\right)a^n=\sum _{j=0}^n\left(\begin{array}{c}n\\ j\end{array}\right)x^{n-j}{a}^j$

Step 2. Explanation

$\Rightarrow {\left(x-2\right)}^6=\left(\begin{array}{c}6\\ 0\end{array}\right){\left(-2\right)}^0{\left(x\right)}^6+\left(\begin{array}{c}6\\ 1\end{array}\right){\left(-2\right)}^1{\left(x\right)}^5+\left(\begin{array}{c}6\\ 2\end{array}\right){\left(-2\right)}^2{\left(x\right)}^4+\left(\begin{array}{c}6\\ 3\end{array}\right){\left(-2\right)}^3{\left(x\right)}^3+\left(\begin{array}{c}6\\ 4\end{array}\right){\left(-2\right)}^4{\left(x\right)}^2+\left(\begin{array}{c}6\\ 5\end{array}\right){\left(-2\right)}^5{\left(x\right)}^1+\left(\begin{array}{c}6\\ 6\end{array}\right){\left(-2\right)}^6{\left(x\right)}^0$

$\Rightarrow {\left(x-2\right)}^6=\frac{6!}{0!\left(6-0\right)!}{x}^6+\frac{6!}{1!\left(6-1\right)!}\left(-2\right)x^5+\frac{6!}{2!\left(6-2\right)!}\left(4\right)x^4+\frac{6!}{3!\left(6-3\right)!}\left(-8\right)x^3+\frac{6!}{4!\left(6-4\right)!}\left(16\right)x^2+\frac{6!}{5!\left(6-5\right)!}\left(-32\right)x^1+\frac{6!}{6!\left(6-6\right)!}\left(64\right)$
$\Rightarrow {\left(x-2\right)}^6=x^6-12x^5+60x^4-160x^3+240x^2-192x+64$