Question

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

Answer 1

Hint: In order to determine the expansion of the above function using binomial theorem, compare it with the form ${\left( {x + y} \right)^n}$ to find out the value of $x$ and $y$ and then use the binomial theorem ${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $ by putting the value of $x$ and $y$ and expand the summation. Simplifying the equation, you will get your required result.

Complete step by step answer:
We are given a function ${\left( {5 - 3y} \right)^3}$, and we have to expand this function with the help of binomial theorem.
Let’s first understand why we are using binomial theorem. The reason is pretty simple, because the power of the function is large. When the power of any function increases, the expansion becomes really very tough and lengthy.
In such situations, binomial theorems come into play which provide a simple formula to expand any function of the form ${\left( {x + y} \right)^n}$ with the use of $^n{C_r}$.
According to the Binomial Theorem:
${\left( {x + y} \right)^n} = \sum\limits_{r = 0}^n {^n{C_r}{x^{n - r}}{y^r}} $
Where, $^n{C_r} = \dfrac{{n!}}{{r!\left( {n - r} \right)!}}$, $n \in N,x,y \in R$
If we expand the summation, we get
${\left( {x + y} \right)^n}{ = ^n}{C_0}{x^n}{y^0}{ + ^n}{C_1}{x^{n - 1}}{y^1}{ + ^n}{C_2}{x^{n - 2}}{y^2}{ + ^n}{C_3}{x^{n - 3}}{y^3} + .....{ + ^n}{C_n}{x^0}{y^n}$
According to our question, we have to expand ${\left( {5 - 3y} \right)^3}$, so if we compare this with ${\left( {x + y} \right)^n}$
We get $n = 3$, $x = 5$ and $y = - 3y$
Now, putting these values in the binomial theorem, we get:
${\left( {5 - 3y} \right)^3} = \sum\limits_{r = 0}^3 {^3{C_r}{{\left( 5 \right)}^{3 - r}}{{\left( { - 3y} \right)}^r}} $
Expanding the summation, we get:
\[{\left( {5 - 3y} \right)^3}{ = ^3}{C_0}{\left( 5 \right)^{3 - 0}}{\left( { - 3y} \right)^0}{ + ^3}{C_1}{\left( 5 \right)^{3 - 1}}{\left( { - 3y} \right)^1}{ + ^3}{C_2}{\left( 5 \right)^{3 - 2}}{\left( { - 3y} \right)^2}{ + ^3}{C_3}{\left( 5 \right)^{3 - 3}}{\left( { - 3y} \right)^3}\]
Simplifying it further, we get:
\[
  {\left( {5 - 3y} \right)^3}{ = ^3}{C_0}{\left( 5 \right)^3}\left( 1 \right){ + ^3}{C_1}{\left( 5 \right)^2}\left( { - 3y} \right){ + ^3}{C_2}\left( 5 \right){\left( { - 3y} \right)^2}{ + ^3}{C_3}\left( 1 \right){\left( { - 3y} \right)^3} \\
  {\left( {5 - 3y} \right)^3}{ = ^3}{C_0}\left( {125} \right)\left( 1 \right){ + ^3}{C_1}\left( {25} \right)\left( { - 3y} \right){ + ^3}{C_2}\left( 5 \right)\left( {9{y^2}} \right){ + ^3}{C_3}\left( 1 \right)\left( { - 27{y^3}} \right) \\
    \\
 \]
Now, as we know that:
$
  ^3{C_0} = \dfrac{{3!}}{{0!3!}} = 1 \\
  ^3{C_1} = \dfrac{{3!}}{{1!2!}} = \dfrac{{3 \times 2 \times 1}}{{1 \times 2 \times 1}} = 3 \\
  ^3{C_2} = \dfrac{{3!}}{{2!1!}} = \dfrac{{3 \times 2 \times 1}}{{2 \times 1 \times 1}} = 3 \\
  ^3{C_3} = \dfrac{{3!}}{{3!0!}} = 1 \\
 $
Therefore, putting these values in our obtained equation, we will get:
\[
  {\left( {5 - 3y} \right)^3} = \left( 1 \right)\left( {125} \right) + \left( 3 \right)\left( {25} \right)\left( { - 3y} \right) + \left( 3 \right)\left( 5 \right)\left( {9{y^2}} \right) + \left( 1 \right)\left( 1 \right)\left( { - 27{y^3}} \right) \\
  {\left( {5 - 3y} \right)^3} = 125 - 225y + 135{y^2} - 27{y^3} \\
 \]

Hence, the binomial expansion of the function ${\left( {5 - 3y} \right)^3}$ is
\[{\left( {5 - 3y} \right)^3} = 125 - 225y + 135{y^2} - 27{y^3}\].

Note: We must know that $^n{C_r}{ = ^n}{C_{n - r}}$. Also, that factorials of proper fraction or negative integers are not defined. Factorials are only defined for whole numbers. Meaning of Zero factorial is senseless to define as the product of integers from one to zero. So, we define it as $0! = 1$. Also, do not forget to cross-check your answer at least once as it may contain calculation mistakes.