Question

Expand using binomial theorem ${\left( {2x + 3y} \right)^3}$

Answer 1

Hint: The given question requires us to find the cube of an algebraic binomial expression. We can find the binomial expansion of the given binomial expression 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 answer:
So, we are required to expand the expression of ${\left( {2x + 3y} \right)^3}$.
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 ${\left( {2x + 3y} \right)^3}$ is \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} \] .
Now, we have to expand the expression \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} \] and we are done with the binomial expansion of ${\left( {2x + 3y} \right)^3}$ .
\[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} = \left( {^3{C_0}} \right){\left( {2x} \right)^3}{\left( {3y} \right)^0} + \left( {^3{C_1}} \right){\left( {2x} \right)^2}{\left( {3y} \right)^1} + \left( {^3{C_2}} \right){\left( {2x} \right)^1}{\left( {3y} \right)^2} + \left( {^3{C_3}} \right){\left( {2x} \right)^0}{\left( {3y} \right)^3}\]
Substituting values of combination formula,
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} = \left( 1 \right){\left( {2x} \right)^3}{\left( {3y} \right)^0} + 3{\left( {2x} \right)^2}{\left( {3y} \right)^1} + 3{\left( {2x} \right)^1}{\left( {3y} \right)^2} + \left( 1 \right){\left( {2x} \right)^0}{\left( {3y} \right)^3}\]
We know that any number raised to power zero is equal to one. So, we get,
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} = {\left( {2x} \right)^3} + 3{\left( {2x} \right)^2}{\left( {3y} \right)^1} + 3{\left( {2x} \right)^1}{\left( {3y} \right)^2} + {\left( {3y} \right)^3}\]
Computing the powers of brackets in the expression above, we get,
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} = 8{x^3} + 3\left( {4{x^2}} \right)\left( {3y} \right) + 3\left( {2x} \right)\left( {9{y^2}} \right) + \left( {27{y^3}} \right)\]
Opening brackets and doing calculations, we get
\[ \Rightarrow \sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} = 8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}\]
Hence, the value of \[\sum\nolimits_{r = 0}^3 {\left( {^3{C_r}} \right){{\left( {2x} \right)}^{3 - r}}{{\left( {3y} \right)}^r}} \] is \[\left( {8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}} \right)\] .
So, Binomial expansion of ${\left( {2x + 3y} \right)^3}$ is \[\left( {8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}} \right)\] .
So, the correct answer is “\[\left( {8{x^3} + 36{x^2}y + 54x{y^2} + 27{y^3}} \right)\]”.

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.