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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 11 of 36
Marks: +1, -0
Find (a+b)4−(a−b)4(a+b)^4-(a-b)^4. Hence, evaluate (3+2)4−(3−2)4(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4
Solution:  
By binomial theorem, we have
(a+b)4−(a−b)4(a+b)^4-(a-b)^4 = [(40)a4+(41)a3b+(42)a2b2\binom{4}{0}a^4+\binom{4}{1}a^3b+\binom{4}{2}a^2b^2 + (43)ab3+(44)b4\binom{4}{3}ab^3+\binom{4}{4}b^4] - [(40)a4−(41)a3b+(42)a2b2\binom{4}{0}a^4-\binom{4}{1}a^3b+\binom{4}{2}a^2b^2 - (43)ab3+(44)b4\binom{4}{3}ab^3+\binom{4}{4}b^4]
= a4+4a3b+6a2b2+4ab3a^4+4a^3b+6a^2b^2+4ab^3 + b4−a4+4a3b−6a2b2b^4-a^4+4a^3b-6a^2b^2 + 4ab3−b44ab^3-b^4
∴ (a+b)4−(a−b)4(a+b)^4-(a-b)^4 = 8a3b+8ab38a^3b+8ab^3 = 8ab(a2+b2)8ab(a^2+b^2) ... (i)
Substituting a = 3\sqrt{3} and b = 2\sqrt{2} , in (i) we get
(3+2)4−(3−2)4(\sqrt{3}+\sqrt{2})^4-(\sqrt{3}-\sqrt{2})^4 = 832[(3)2+(2)2]8\sqrt{3}\sqrt{2}[(\sqrt{3})^2+(\sqrt{2})^2] = 86[3+2]8\sqrt{6}[3+2] = 40640\sqrt{6}
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