logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
0 votes

Find $ (a + b)^4 - (a - b)^4.$ Hence, evaluate $( \sqrt3 + \sqrt2)^4 - (\sqrt 3 - \sqrt 2)^4$.

$\begin{array}{1 1} 40 \sqrt 6 \\ 20 \sqrt 6 \\ 40 \sqrt 3 \\ 20 \sqrt 3 \end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
Toolbox:
  • Express the given number as the sum or difference of two numbers whose powers are easier to evaluate. Then use binomial theorem as follows:
  • $(a+b)^n = \large \sum \limits_{k=0}^{n}\; $$^n \large C$$_k \; a^{n-k}b^k =\; ^n \large C$$_0\;a^nb^0 +\; ^n \large C$$_1\;a^{n-1}b^1+.... ^n\large C$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
We can now use Binomial Theorem to evaluate $ (a + b)^4 – (a – b)^4$
$(a+b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+....$
$(a-b)^4 = ^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1+....$
$ (a + b)^4 – (a – b)^4 = ^4\large C$$_0\;a^nb^0 + ^4\large C$$_1\;a^{4-1}b^1+.... ^4\large C$$_4\;a^0b^4 - (^4\large C$$_0\;a^nb^0 - ^4\large C$$_1\;a^{4-1}b^1 + ^4\large C$$_4\;a^{4-4}b^4$)
$\qquad = 2 (\;^4\large C$$_1\;a^3b+\;^4\large C$$_3\;ab^3)$
$\qquad = 2 (4a^3b+4ab^3)$
$\qquad = 8ab(a^2+b^2)$
Therefore, $( \sqrt3 + \sqrt2)^4– (\sqrt 3 –\sqrt 2)^4 = 8 \sqrt 3 \sqrt 2 (\sqrt 3^2 + \sqrt 2^2)$
$\qquad = 8 \times \sqrt 6 \times 5 = 40 \sqrt 6 \approx 97.98$

 

answered Apr 10, 2014 by balaji.thirumalai
edited Apr 10, 2014 by balaji.thirumalai
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...