Question

Find the expansion of ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$ using binomial theorem.

Answer 1

Hint: So we have to find the expansion of ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$ using binomial theorem. Take a value $a$ and $b$ from ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$. Use binomial theorem. You will get the answer.

Complete step-by-step answer:
According to the binomial theorem, the ${{(r+1)}^{th}}$ term in the expansion of ${{(a+b)}^{n}}$ is,
\[{{T}_{r+1}}={}^{n}{{C}_{r}}{{a}^{n-r}}{{b}^{r}}\]
The above term is a general term or ${{(r+1)}^{th}}$ term. The total number of terms in the binomial expansion ${{(a+b)}^{n}}$is$(n+1)$, i.e. one more than the exponent $n$.

Binomial theorem states that for any positive integer $n$, the $n$ power of the sum of two numbers a and b may be expressed as the sum of $(n+1)$ terms of the form.

The final expression follows from the previous one by the symmetry of $a$ and $b$ in the first expression, and by comparison, it follows that the sequence of binomial coefficients in the formula is symmetrical.

A simple variant of the binomial formula is obtained by substituting $1$ for $b$ so that it involves only a single variable.
In the Binomial expression, we have
\[{{(a+b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}+{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}+{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
So the coefficients ${}^{n}{{C}_{0}},{}^{n}{{C}_{1}},............,{}^{n}{{C}_{n}}$ are known as binomial or combinatorial coefficients.

You can see them ${}^{n}{{C}_{r}}$ being used here which is the binomial coefficient. The sum of the binomial coefficients will be ${{2}^{n}}$ because, we know that,
$\sum\nolimits_{r=0}^{n}{\left( {}^{n}{{C}_{r}} \right)}={{2}^{n}}$
Thus, the sum of all the odd binomial coefficients is equal to the sum of all the even binomial coefficients and each is equal to ${{2}^{n-1}}$.

The middle term depends upon the value of $n$,
It $n$ is even: then the total number of terms in the expansion of${{(a+b)}^{n}}$ is $n+1$ (odd).
It $n$ is odd: then the total number of terms in the expansion of ${{(a+b)}^{n}}$ is $n+1$ (even).

It $n$is a positive integer,
\[{{(a-b)}^{n}}=={}^{n}{{C}_{0}}{{a}^{n}}{{\left( b \right)}^{0}}-{}^{n}{{C}_{1}}{{a}^{n-1}}{{\left( b \right)}^{1}}+{}^{n}{{C}_{2}}{{a}^{n-2}}{{\left( b \right)}^{2}}-{}^{n}{{C}_{3}}{{a}^{n-3}}{{\left( b \right)}^{3}}+...........+{}^{n}{{C}_{n}}{{a}^{0}}{{\left( b \right)}^{n}}\]
For binomial expansion first, let's do a small pairing inside the bracket.
So now let$a=3{{x}^{2}}$and$b=a(2x-3a)$.
Now let's expand this as is normally done for two-digit expansion.
${{\left[ 3{{x}^{2}}-a(2x-3a) \right]}^{3}}={{(3{{x}^{2}})}^{3}}-3{{(3{{x}^{2}})}^{2}}\times a(2x-3a)+3(3{{x}^{2}})\times {{a}^{2}}{{(2x-3a)}^{2}}-{{a}^{3}}{{(2x-3a)}^{3}}$
So simplifying in a simple manner we get,
$\begin{align}
  & {{\left[ 3{{x}^{2}}-a(2x-3a) \right]}^{3}}=27{{x}^{6}}-3(9{{x}^{4}})(2ax-3{{a}^{2}})+9{{a}^{2}}{{x}^{2}}(4{{x}^{2}}+9{{a}^{2}}-12ax)-{{a}^{3}}(8{{x}^{3}}-27{{a}^{3}}-3.4{{x}^{2}}.3a+3.9{{a}^{2}}.2x) \\
 & =27{{x}^{6}}-27{{x}^{4}}(2ax-3{{a}^{2}})+36{{a}^{2}}{{x}^{4}}+81{{a}^{4}}{{x}^{2}}-108{{a}^{3}}{{x}^{3}}-8{{a}^{3}}{{x}^{3}}+27{{a}^{6}}+36{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x \\
 & =27{{x}^{6}}-54a{{x}^{5}}+81{{a}^{2}}{{x}^{4}}+36{{a}^{2}}{{x}^{4}}+117{{a}^{4}}{{x}^{2}}-116{{a}^{3}}{{x}^{3}}+27{{a}^{6}}-54{{a}^{5}}x \\
 & =27{{x}^{6}}-54a{{x}^{5}}+117{{a}^{2}}{{x}^{4}}-116{{a}^{3}}{{x}^{3}}+117{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}} \\
\end{align}$

The Expansion ${{(3{{x}^{2}}-2ax+3{{a}^{2}})}^{3}}$is$27{{x}^{6}}-54{{x}^{5}}a+117{{a}^{2}}{{x}^{4}}-116{{a}^{3}}{{x}^{ e3}}+117{{a}^{4}}{{x}^{2}}-54{{a}^{5}}x+27{{a}^{6}}$.

Note: Read the question in a careful manner. Don’t jumble within the concepts. You should know what to select as$a$ and $b$. We had assumed $a=3{{x}^{2}}$ and $b=a(2x-3a)$. So you can assume it in your way. But keep in mind there should not be any confusion.