Question

$(a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4$ का मान ज्ञात कीजिए।

Answer 1

Answer:

Step-by-step explanation:

$(a^2 + \sqrt{a^2 - 1})^4 + (a^2 - \sqrt{a^2 - 1})^4\\\\=[^4C_0(a^2)^4+^4C_1(a^2)^3(\sqrt{a^2-1}  )+^4C_2(a^2)^2(\sqrt{a^2-1}  )^2+^4C_3(a^2)^1(\sqrt{a^2-1}  )^3+^4C_4(\sqrt{a^2-1}  )^4]+[^4C_0(a^2)^4-^4C_1(a^2)^3(\sqrt{a^2-1}  )+^4C_2(a^2)^2(\sqrt{a^2-1}  )^2-^4C_3(a^2)(\sqrt{a^2-1}  )^3+^4C_4(\sqrt{a^2-1}  )^4]\\\\=2[^4C_0(a^2)^4+^4C_2(a^2)^2(\sqrt{a^2-1}  )^2+^4C_4(\sqrt{a^2-1}  )^4]$

अन्य पदों की उपेक्षा करने पर

$=2[a^8+6a^4(\sqrt{a^2-1} )^2+(\sqrt{a^2-1} )^4]\\\\=2[a^8+6a^4(a^2-1)+(a^2-1)^2]\\\\=2[a^8+6a^6-6a^4+a^4-2a^2+1]\\\\=2[a^8+6a^6-5a^4-2a^2+1]\\\\=2a^8+12a^6-10a^4-4a^2+2$