Question

How do you use the binomial formula to expand ${{\left( 2x+3 \right)}^{3}}$?

Answer 1

Hint: In this question, we are given an expression in terms of x which we need to expand using the binomial formula. The binomial formula for addition of two numbers having the exponential power as 3 is given by ${{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}$. So we will compare this with given expansion and find a and b so that we can apply this formula and expand the given expression.

Complete step by step solution:
Here we are given the expression as ${{\left( 2x+3 \right)}^{3}}$. We need to expand it using binomial formula. For this let us first understand the binomial expansion. For any two numbers if their sum is to the power of n then the expansion is given as ${{\left( a+b \right)}^{n}}=\sum\limits_{k=0}^{n}{{}^{n}{{C}_{k}}}{{a}^{n-k}}{{b}^{k}}$.
Here we have the degree as 3. So the value of n becomes 3. Hence the formula becomes ${{\left( a+b \right)}^{3}}=\sum\limits_{k=0}^{3}{{}^{3}{{C}_{k}}}{{a}^{3-k}}{{b}^{k}}\Rightarrow {}^{3}{{C}_{0}}{{a}^{3}}{{b}^{0}}+{}^{3}{{C}_{1}}{{a}^{3-1}}{{b}^{1}}+{}^{3}{{C}_{2}}{{a}^{3-2}}{{b}^{2}}+{}^{3}{{C}_{3}}{{a}^{3-3}}{{b}^{3}}$.
We know that ${{x}^{0}}=1$ where x is any number. So, above formula becomes ${{\left( a+b \right)}^{3}}={}^{3}{{C}_{0}}{{a}^{3}}+{}^{3}{{C}_{1}}{{a}^{2}}b+{}^{3}{{C}_{2}}a{{b}^{2}}+{}^{3}{{C}_{3}}{{b}^{3}}$.
Now let us calculate the combination. We know that $^{n}{{C}_{r}}=\dfrac{n!}{r!\left( n-r \right)!}$. So let us use this and find all the coefficient, we get ${{\left( a+b \right)}^{3}}=\dfrac{3!}{0!3!}{{a}^{3}}+\dfrac{3!}{1!2!}{{a}^{2}}b+\dfrac{3!}{2!1!}a{{b}^{2}}+\dfrac{3!}{3!0!}{{b}^{3}}$.
We know that 0! is equal to 1 and 3! can be written as $3\times 2\times 1$ and 2! can be written as $2\times 1$ and 1! is equal to 1 so using all this we get \[{{\left( a+b \right)}^{3}}={{a}^{3}}+\dfrac{3\times 2\times 1}{2\times 1}{{a}^{2}}b+\dfrac{3\times 2\times 1}{2\times 1}a{{b}^{2}}+{{b}^{3}}\Rightarrow {{\left( a+b \right)}^{3}}={{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\].
So the binomial formula for power of 3 become \[{{a}^{3}}+3{{a}^{2}}b+3a{{b}^{2}}+{{b}^{3}}\].
Now let us compare \[{{\left( a+b \right)}^{3}}\] with \[{{\left( 2x+3 \right)}^{3}}\] to get the expansion. We see that a = 2x and b = 3. So using the binomial formula we get \[{{\left( 2x+3 \right)}^{3}}={{\left( 2x \right)}^{3}}+3{{\left( 2x \right)}^{2}}\left( 3 \right)+3\left( 2x \right){{\left( 3 \right)}^{2}}+{{\left( 3 \right)}^{3}}\].
We know that ${{\left( ab \right)}^{n}}$ can be written as ${{a}^{n}}{{b}^{n}}$. So using this and solving the ${{3}^{3}}$ as $3\times 3\times 3=27$ we get \[{{\left( 2x+3 \right)}^{3}}={{\left( 2 \right)}^{3}}{{x}^{3}}+9{{\left( 2 \right)}^{2}}{{x}^{2}}+3\times 9\times 2x+27\].
Solving all the required square, cubes and multiplication we get \[{{\left( 2x+3 \right)}^{3}}=8{{x}^{3}}+36{{x}^{2}}+54x+27\].
Hence this is the required expansion of \[{{\left( 2x+3 \right)}^{3}}\].

Note: Students should try to keep in mind the binomial formula for some degree such as 2 and 3. Student must take care while calculating the combination i.e. $^{n}{{C}_{r}}$. Note that 0! is always equal to 1. Make sure to take cube of 2 and x both in ${{\left( 2x \right)}^{3}}$.