Question

Use the Binomial Theorem to expand${\left(y-1\right)}^6$.

Short answer

The expansion of the given binomial is, $y^6-6y^5+15y^4-20y^3+15y^2-6y+1$

Step-by-step solution

Step 1. Given information. 

The given expression is, ${\left(y-1\right)}^6$

Step 2. Substitute the values and simplify.  

Substitute in the values into the binomial expansion,

$$\begin{gathered} a=y \\ b=-1 \\ n=6 \end{gathered}$$

Now

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

Simplify the exponents.

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

Evaluate the coefficients. Remember,

$\left(\begin{array}{c}n\\ 1\end{array}\right)=n,\left(\begin{array}{c}n\\ n\end{array}\right)=1,\left(\begin{array}{c}n\\ 0\end{array}\right)=1$

Simplify the exponents,

${\left(y-1\right)}^6=1y^6-6y^5+\frac{6!}{2!4!}{y}^4-\frac{6!}{3!3!}{y}^3+\frac{6!}{4!2!}{y}^2-\frac{6!}{5!1!}y+1$

Step 3. Conclusion.

Therefore, the expansion of the given expression is ${\left(y-1\right)}^6=y^6-6y^5+15y^4-20y^3+15y^2-6y+1$