Chapter 6.1, Problem 28E

Solution

To write: the binomial expansion using the binomial theorem.

  ${\left(w^3-3\right)}^4=w^{12}-12w^9+54w^6-108w^3+81$ .

Given information:

The given expression is:

  ${\left(w^3-3\right)}^4$

Formula Used:

Combinations of n Objects Taken r at a Time

The number of combinations of r objects taken from a group of n distinct objects is denoted by $C{}_r$ and is given by the formula:

  $C{}_r=\frac{n!}{\left(n-r\right)!\cdot r!}$ .

  $\begin{array}{l}C{}_0=1\\ C{}_n=1\end{array}$

Theorem Used:

Binomial Theorem

For any positive integer n , the binomial expansion of ${\left(a+b\right)}^n$ is:

  ${\left(a+b\right)}^n=C{}_0{a}^n{b}^0+C{}_1{a}^{n-1}{b}^1+C{}_2{a}^{n-2}{b}^2+\cdots +C{}_n{a}^0{b}^n$

Here, each term in the expansion of ${\left(a+b\right)}^n$ has the form $C{}_r{a}^{n-r}{b}^r$ where r s an integer from 0 to n .

Explanation:

By using the binomial theorem, the given expression ${\left(w^3-3\right)}^4$ is written in expanded form as:

  $\begin{array}{c}{\left(w^3-3\right)}^4={\left(w^3+\left(-3\right)\right)}^4\\ =C{}_0{\left(w^3\right)}^4{\left(-3\right)}^0+C{}_1{\left(w^3\right)}^3{\left(-3\right)}^1+C{}_2{\left(w^3\right)}^2{\left(-3\right)}^2+C{}_3{\left(w^3\right)}^1{\left(-3\right)}^3+C{}_4{\left(w^3\right)}^0{\left(-3\right)}^4\\ =\left(\frac{4!}{\left(4-0\right)!0!}\right)\left(w^{3\cdot 4}\right)\left(1\right)+\left(\frac{4!}{\left(4-1\right)!1!}\right)\left(w^{3\cdot 3}\right)\left(-3\right)+\left(\frac{4!}{\left(4-2\right)!2!}\right)\left(w^{3\cdot 2}\right)\left(9\right)\\ +\left(\frac{4!}{\left(4-3\right)!3!}\right)\left(w^{3\cdot 1}\right)\left(-27\right)+\left(\frac{4!}{\left(4-4\right)!4!}\right)\left(w^{3\cdot 0}\right)\left(81\right)\\ =1\left(w^{12}\right)\left(1\right)+\left(4\right)\left(w^9\right)\left(-3\right)+\left(6\right)\left(w^6\right)\left(9\right)+\left(4\right)\left(w^3\right)\left(-27\right)+\left(1\right)\left(1\right)\left(81\right)\\ =w^{12}-12w^9+54w^6-108w^3+81\end{array}$