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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 31 of 36
Marks: +1, -0
Evaluate (√3+√2)6−(√3−√2)6(√3+√2)^6-(√3-√2)^6
Solution:  
We have (√3+√2)6(√3+√2)^6 =
 ↖6C0(√3)6+ ↖6C1(√3)5(√2)+ ↖6C2(√3)4(√2)2+ ↖6C3(√3)3(√2)3+ ↖6C4(√3)2(√2)4+ ↖6C5(√3)(√2)5+ ↖6C6(√2)6\,↖{6}C_0(√3)^6 + \,↖{6}C_1(√3)^5(√2) +\,↖{6}C_2 (√3)^4 (√2)^2 + \,↖{6}C_3 (√3)^3 (√2)^3 + \,↖{6}C_4 (√3)^2 (√2)^4 + \,↖{6}C_5 (√3)(√2)^5 + \,↖{6}C_6 (√2)^6
... (i)
and (√3−√2)6(√3-√2)^6 =
 ↖6C0(√3)6− ↖6C1(√3)5(√2)+ ↖6C2(√3)4(√2)2− ↖6C3(√3)3(√2)3+ ↖6C4(√3)2(√2)4− ↖6C5(√3)(√2)5+ ↖6C6(√2)6\,↖{6}C_0(√3)^6 - \,↖{6}C_1(√3)^5(√2) + \,↖{6}C_2(√3)^4(√2)^2 - \,↖{6}C_3 (√3)^3 (√2)^3 + \,↖{6}C_4 (√3)^2 (√2)^4 - \,↖{6}C_5 (√3)(√2)^5 + \,↖{6}C_6 (√2)^6
... (ii)
Subtracting (ii) from (i), we get (√3+√2)6−(√3−√2)6(√3+√2)^6-(√3-√2)^6
=
2[ ↖6C1(√3)5(√2)+ ↖6C3(√2)3(√2)3+ ↖6C5(√3)(√2)5]2[\,↖{6}C_1(√3)^5 (√2) + \,↖{6}C_3(√2)^3(√2)^3+\,↖{6}C_5(√3)(√2)^5]
=
2[6(9√3)(√2)+20(3√3)(2√2)+6(√3)(4√2)]2[6(9√3)(√2)+20(3√3)(2√2)+6(√3)(4√2)]
= 2 [54√6+120√6+24√6][54√6 + 120√6 + 24√6] = 2 (198√6)(198√6) = 396 √6√6
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