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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Using Binomial Theorem, evaluate $96^3$.

$\begin{array}{1 1}884736 \\ 19876\\ 25634 \\ 574326 \end{array} $

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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$$_n\;a^0b^n$, where $b^0 = 1 = a^{n-n}$
$96 = 100 - 4 \rightarrow 96^3 = (100-4)^3$. We can now use Binomial Theorem to evaluate this.
$(100-4)^3 = ^3 \large C$$_0\;100^{3-1}4^0 -.^3 \large C$$_1\;100^{3-2}4^1 +.^3 \large C$$_2\;100^{3-2}4^2 -.^3 \large C$$_3\;100^{3-3}4^3 $
$\qquad = 100^3 - 3\;100^2\;4 + 3\;(100) (4^2) - 4^3$
$\qquad = 1000000 - 120000 + 4800 - 64 = 884736$
answered Apr 10, 2014 by balaji.thirumalai
 

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