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Home  >>  CBSE XI  >>  Math  >>  Binomial Theorem
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Expand each of the expressions:$(96)^3$

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  • $(a-b)^n=nC_0a^n-nC_1a^{n-1}b+nC_2a^{n-2}b^2.......+(-1)^nnC_ra^{n-r}b^r+......nC_n(-b)^n$
$(96)^3=(100-4)^3$
$\Rightarrow 3C_0(100)^3(-4)^0+3C_1(100)^2(-4)+3C_2(100)^1(-4)^2+3C_3(100)^0(-4)^3$
$\Rightarrow (100)^3+\large\frac{3}{1}$$(100)^2(-4)+\large\frac{3}{1}$$\times 100\times 16+(-64)$
$\Rightarrow 1000000-12\times 10000+48\times 100-64$
$\Rightarrow 1000000-120000+4800-64$
$\Rightarrow 884736$
Hence (A) is the correct answer.
answered Jun 20, 2014 by sreemathi.v
 
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