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NCERT Class XI Mathematics - Binomial Theorem - Solutions

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Question : 32 of 36
Marks: +1, -0
Find the value of
(a2+a2−1)4+(a2−a2−1)4(a^2+\sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4
Solution:  
We have, (a2+a2−1)4(a^2+\sqrt{a^2-1})^4 =
4C0(a2)4+4C1(a2)3(a2−1)+4C2(a2)2(a2−1)2+4C3(a2)(a2−1)3+4C4(a2−1)4{}^{4}C_{0}(a^2)^4 + {}^{4}C_{1} (a^2)^3(\sqrt{a^2-1}) + {}^{4}C_{2}(a^2)^2(\sqrt{a^2-1})^2 + {}^{4}C_{3} (a^2)(\sqrt{a^2-1})^3+{}^{4}C_{4} (\sqrt{a^2-1})^4
... (i)
and (a2−a2−1)4(a^2 - \sqrt{a^2-1})^4 =
4C0(a2)4−4C1(a2)3(a2−1)+4C2(a2)2(a2−1)2−4C3(a2)(a2−1)3+4C4(a2−1)4{}^{4}C_{0}(a^2)^4 - {}^{4}C_{1} (a^2)^3(\sqrt{a^2-1}) +{}^{4}C_{2}(a^2)^2(\sqrt{a^2-1})^2 - {}^{4}C_{3} (a^2) (\sqrt{a^2-1})^3+{}^{4}C_{4}(\sqrt{a^2-1})^4
... (ii)
Adding (i) and (ii), we get
(a2+a2−1)4+(a2−a2−1)4=2[4C0(a2)4+4C2(a2)2(a2−1)2+4C4(a2−1)4](a^2+\sqrt{a^2-1})^4 + (a^2 - \sqrt{a^2-1})^4 = 2[{}^{4}C_{0} (a^2)^4 + {}^{4}C_{2}(a^2)^2(\sqrt{a^2-1})^2+{}^{4}C_{4}(\sqrt{a^2-1})^4]
= 2[a8+6a4(a2−1)+(a2−1)2]2[a^8+6a^4(a^2-1)+(a^2-1)^2] = 2[a8+6a6−6a4+a4−2a2+1]2[a^8+6a^6-6a^4+a^4-2a^2+1]
= 2a8+12a6−10a4−4a2+22a^8+12a^6-10a^4-4a^2+2
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