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Binomial Theorem — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsBinomial Theorem3 Marks
Question
Find $(a + b)^4 - (a - b)^4.$ Hence, evaluate ${(\sqrt 3 + \sqrt 2 )^4} - {(\sqrt 3 - \sqrt 2 )^4}$
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Answer

$(a + b)^4 = {[^4}{C_0}{a^4}{ + ^4}{C_1}{a^3}b{ + ^4}{C_2}{a^2}{b^2}$${ + ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$
and ${\left(a-b\right)^4{\;=\lbrack^4}C_0a^4-^4C_1a^3b+^4C_2a^2{b^2}}$${ - ^4}{C_3}a{b^3}{ + ^4}{C_4}{b^4}]$
${\therefore\;(a+b)^4\;-\left(a-b\right)^4\;=}2\left[{}^4C_1a^3b\;+^4C_3ab^3\right]$
$=2\left[4a^3b\;+4ab^3\right]\;=\;8ab\left[a^2+b^2\right]$
$\therefore\left(\sqrt3\;+\sqrt2\right)^4\;-\;\left(\sqrt3\;-\sqrt2\right)^{4\;}=\;8.\sqrt3.\sqrt2\left[\left(\sqrt3\right)^2+\left(\sqrt2\right)^2\right]$
$=\;8.\sqrt3.\sqrt2\left[3+2\right]\;=\;40.\sqrt3.\sqrt2\;=\;40\sqrt6$

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