Question

How do you use the binomial theorem to expand and simplify the expression ${(5-3\mathrm{y})}^3$?

Answer 1

Step-1 : Explain the Binomial theorem:

Any polynomial with two terms (or) an algebraic expression with two terms is called a binomial.

The binomial theorem helps to find the powers of a binomial that cannot be expressed by using the algebraic identities. The algebraic expression for the binomial theorem formula is : ${(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}=\sum _{\mathrm{r}=0}^{\mathrm{n}}{}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}{\mathrm{a}}^{\mathrm{n}-\mathrm{r}}{\mathrm{b}}^{\mathrm{r}}$

where, $\mathrm{a},\mathrm{b}$ - real numbers

$\mathrm{n}$ - positive integer

${}^{\mathrm{n}}\mathrm{C}_{\mathrm{r}}=\frac{\mathrm{n}!}{\mathrm{r}!(\mathrm{n}-\mathrm{r})!}$is binomial coefficient.

Step- 2 : Simplify the expression:

The given polynomial expression ${(5-3\mathrm{y})}^3$ is of the form ${(\mathrm{a}+\mathrm{b})}^{\mathrm{n}}$.

where $\mathrm{a}=5$, $\mathrm{b}=-3y$ and $\mathrm{n}=3$.

$$\begin{gathered} {(5-3\mathrm{y})}^3=\sum _{\mathrm{r}=0}^3{}^3\mathrm{C}_{\mathrm{r}}{5}^{3-\mathrm{r}}{(-3\mathrm{y})}^{\mathrm{r}} \\ ={}^3\mathrm{C}_0{5}^{3-0}{(-3\mathrm{y})}^0+{}^3\mathrm{C}_1{5}^{3-1}{(-3\mathrm{y})}^1+{}^3\mathrm{C}_2{5}^{3-2}{(-3\mathrm{y})}^2+{}^3\mathrm{C}_3{5}^{3-3}{(-3\mathrm{y})}^3 \\ =\frac{3!}{0!(3-0)!}{5}^3{(-3\mathrm{y})}^0+\frac{3!}{1!(3-1)!}{5}^2{(-3\mathrm{y})}^1+\frac{3!}{2!(3-2)!}{5}^1{(-3\mathrm{y})}^2+\frac{3!}{3!(3-3)!}{5}^0{(-3\mathrm{y})}^3 \\ =\frac{3!}{0!(3!)}{5}^3{(-3\mathrm{y})}^0+\frac{3!}{1!(2!)}{5}^2{(-3\mathrm{y})}^1+\frac{3!}{2!(1!)}{5}^1{(-3\mathrm{y})}^2+\frac{3!}{3!(0!)}{5}^0{(-3\mathrm{y})}^3 \\ =1\times 5^3{(-3\mathrm{y})}^0+\frac{3\times 2\times 1}{2\times 1}{5}^2{(-3\mathrm{y})}^1+\frac{3\times 2\times 1}{2\times 1}{5}^1{(-3\mathrm{y})}^2+\frac{3\times 2\times 1}{3\times 2\times 1}{5}^0{(-3\mathrm{y})}^3 \\ =1\times 5^3{(-3\mathrm{y})}^0+3\times 5^2{(-3\mathrm{y})}^1+3\times 5^1{(-3\mathrm{y})}^2+1\times 5^0{(-3\mathrm{y})}^3 \\ =125-225\mathrm{y}+135{\mathrm{y}}^2-27{\mathrm{y}}^3 \end{gathered}$$

Hence, the binomial expansion of ${(5-3\mathrm{y})}^3$ is $125-225\mathrm{y}+135{\mathrm{y}}^2-27{\mathrm{y}}^3$.