Question

Expand \[{\left( {4x + 3} \right)^3}\].

Answer 1

Hint: The given question requires us to find the binomial expansion of the given binomial expression. We can find the required binomial expansion by using the Binomial theorem. Binomial theorem helps us to expand the powers of binomial expressions easily and can be used to solve the given problem. We must know the formulae of combinations and factorials for solving the given question using the binomial theorem.

Complete step by step solution:
We have to find the binomial expansion of \[{\left( {4x + 3} \right)^3}\] . So, using the binomial theorem, the binomial expansion of \[{\left( {x + y} \right)^n}\] is $\sum\nolimits_{r = 0}^n {\left( {^n{C_r}} \right){{\left( x \right)}^{n - r}}{{\left( y \right)}^r}} $ .
So, the binomial expansion of \[{\left( {4x + 3} \right)^3}\] is \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} \] .
Now, we have to expand the expression \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} \] and we are done with the binomial expansion of \[{\left( {4x + 3} \right)^3}\].
\[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = \left( {^3{C_0}} \right){\left( {4x} \right)^3}{\left( 3 \right)^0} + \left( {^3{C_1}} \right){\left( {4x} \right)^2}{\left( 3 \right)^1} + \left( {^3{C_2}} \right){\left( {4x} \right)^1}{\left( 3 \right)^2} + \left( {^3{C_3}} \right){\left( {4x} \right)^0}{\left( 3 \right)^3}\]
Substituting values of combination formulae and simplifying calculations, we get
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = \left( 1 \right){\left( {4x} \right)^3}{\left( 3 \right)^0} + 3{\left( {4x} \right)^2}{\left( 3 \right)^1} + 3{\left( {4x} \right)^1}{\left( 3 \right)^2} + \left( 1 \right){\left( {4x} \right)^0}{\left( 3 \right)^3}\]
Substituting the values of powers of three, we get,
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = \left( 1 \right){\left( {4x} \right)^3}\left( 1 \right) + 3{\left( {4x} \right)^2}\left( 3 \right) + 3{\left( {4x} \right)^1}\left( 9 \right) + \left( 1 \right){\left( {4x} \right)^0}\left( {27} \right)\]
Opening the brackets and simplifying the expression, we get,
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = \left( 1 \right)\left( {64{x^3}} \right)\left( 1 \right) + 3\left( {16{x^2}} \right)\left( 3 \right) + 3\left( {4x} \right)\left( 9 \right) + \left( 1 \right)\left( 1 \right)\left( {27} \right)\]
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = 64{x^3} + 9 \times 16{x^2} + 27 \times 4x + 27\]
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} = 64{x^3} + 144{x^2} + 108x + 27\]
Hence, the value of \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {4x} \right)}^{3 - r}}{{\left( 3 \right)}^r}} \] is \[64{x^3} + 144{x^2} + 108x + 27\] .
Therefore, Binomial expansion of \[{\left( {4x + 3} \right)^3}\] is \[64{x^3} + 144{x^2} + 108x + 27\].

Note:
The given problem can be solved by various methods. The easiest way is to apply the concepts of Binomial theorem as it is very effective in finding the binomial expansion. But the problem can also be solved by actually calculating the cube of the binomial expansion given, though it is tedious to calculate the cube of a binomial polynomial.
We can use the identity ${\left( {a + b} \right)^3} = {a^3} + {b^3} + 3ab\left( {a + b} \right)$.
This familiar formula is actually derived from the binomial theorem.
Here, $a = 4x$ and $b = 3$.
So, we get,
${\left( {4x + 3} \right)^3} = {\left( {4x} \right)^3} + {3^3} + 3\left( {4x} \right)\left( 3 \right)\left( {4x + 3} \right)$
$ \Rightarrow {\left( {4x + 3} \right)^3} = 64{x^3} + 27 + 36x\left( {4x + 3} \right)$
$ \Rightarrow {\left( {4x + 3} \right)^3} = 64{x^3} + 27 + 144{x^2} + 108x$
We got the same answer in both the methods.