Find the expansion of \((3x^2 – 2ax + 3a^2)^3\) using binomial theorem.
Solution and Explanation
Using Binomial Theorem, the given expression \((3x^2-2ax +3a^2)^3\) can be expanded as
\([(3x^2-2ax)+3a^2]^3\)
= \(^3C_0 (3x^2 -2ax)^3 + \space^3C_1(3x^2 - 2ax)^2 (3a^2) +\space ^3C_2 (3x^2 - 2ax) (3a^2)^2 + \space^3C_3 (3a^2)^2\)
\(=(3x^2-2ax)^3+3(9x^4-12ax^3 +4a^2x^2) (3a^2)+3(3x^2-2ax) (9a^4)+27a^6\)
\(=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3 +36a^4x^2+81a^4x^2-54a^5x+27a^6\)
\(=(3x^2-2ax)^3 +81a^2x^4-108a^3x^3+117a^4x^2-54a^5x+27a^6 ...(1)\)
Again by using Binomial Theorem, we obtain
\((3x^2-2ax)^3\)
\(=\space^ 3C_0 (3x^2)^3 - \space^3C_1 (3x^2)^2 (2ax) + \space^3C_2 (3x^2) (2ax)^2 - 3C^3 (2ax)^3 \)
\(=27x^6-3(9x^4) (2ax)+3(3x^2) (4a^2x^2)-8a^3x^3\)
\(=27x^6-54ax^5 +36a^2x^4+-8a^3x^3 ...(2)\)
From (1) and (2), we obtain
\((3x^2-2ax +3a^2)^3\)
\(=27x^6-54ax^5 +36a^2x^4-8a^3x^3 +81a^2x^4-108a^3x^3 +117a^4x^2 - 54a^5x+27a^6\)
\(=27x^6-54ax^5+117a^2x^4+-116a^3x^3 +117a^4x^2 -54a^5 +27a^6\)
Top Questions on Binomial theorem
- The sum of the coefficients of \( x^{2/3} \) and \( x^{-2/5} \) in the binomial expansion of \[ \left( x^{2/3} + \frac{1}{2} x^{-2/5} \right)^9 \] is:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $m$ and $n$ be the coefficients of the seventh and thirteenth terms respectively in the expansion of \[\left( \frac{1}{3}x^{\frac{1}{3}} + \frac{1}{2x^{\frac{2}{3}}} \right)^{18}.\]Then \[\left(\frac{n}{m}\right)^{\frac{1}{3}}\]is:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- If the term independent of \(x\) in the expansion of \[ \left( \sqrt{ax^2} + \frac{1}{2x^3} \right)^{10} \] is 105, then \(a^2\) is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $\alpha = \sum_{r=0}^n (4r^2 + 2r + 1) \binom{n}{r}$ and $\beta = \left( \sum_{r=0}^n \frac{\binom{n}{r}}{r+1} \right) + \frac{1}{n+1}$. If $140 < \frac{2\alpha}{\beta} < 281$, then the value of $n$ is _____.
- JEE Main - 2024
- Mathematics
- Binomial theorem
- Let $0 \leq r \leq n$. If \[^nC_{r+1} : ^nC_r : ^nC_{r-1} = 55 : 35 : 21,\]then $2n + 5r$ is equal to:
- JEE Main - 2024
- Mathematics
- Binomial theorem
Questions Asked in CBSE Class XI exam
- Draw formulas for the first five members of each homologous series beginning with the following compounds.\((a) H–COOH (b) CH_3COCH_3 (c) H–CH=CH_2\)
- CBSE Class XI
- Organic Chemistry- Some Basic Principles and Techniques
- Distinguish between the following pairs of sentences.
(i) He was visibly moved.
(ii) He was visually impaired.- CBSE Class XI
- The adventure
- How, according to you, can peace and liberty be maintained in a state?
- CBSE Class XI
- The tale of melon City
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?
- CBSE Class XI
- Pressure
- Describe the change in hybridisation (if any) of the Al atom in the following reaction.
\(AlCl_3+Cl^- \rightarrow AlCl_4^-\)- CBSE Class XI
- Hybridisation
Concepts Used:
Binomial Expansion Formula
The binomial expansion formula involves binomial coefficients which are of the form
(n/k)(or) nCk and it is calculated using the formula, nCk =n! / [(n - k)! k!]. The binomial expansion formula is also known as the binomial theorem. Here are the binomial expansion formulas.
This binomial expansion formula gives the expansion of (x + y)n where 'n' is a natural number. The expansion of (x + y)n has (n + 1) terms. This formula says:
We have (x + y)n = nC0 xn + nC1 xn-1 . y + nC2 xn-2 . y2 + … + nCn yn
General Term = Tr+1 = nCr xn-r . yr
- General Term in (1 + x)n is nCr xr
- In the binomial expansion of (x + y)n , the rth term from end is (n – r + 2)th .