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  • Find the value of Binomial Theroem Exercise Miscellaneous 6.

Q6. Find the value of \left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4

Answers (1)

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First, lets simplify the expression (x+y)^4-(x-y)^4 using binomial expansion,

(x+y)^4=^4C_0x^4+^4C_1x^3y+^4C_2x^2y^2+^4C_3xy^3+^4C_4y^4

(x+y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4

And

(x-y)^4=^4C_0x^4-^4C_1x^3y+^4C_2x^2y^2-^4C_3xy^3+^4C_4y^4

(x-y)^4=x^4-4x^3y+6x^2y^2-4xy^3+y^4

Now,

(x+y)^4-(x-y)^4=x^4+4x^3y+6x^2y^2+4xy^3+y^4-x^4+4x^3y-6x^2y^2+4xy^3-y^4

(x+y)^4-(x-y)^4=2(x^4+6x^2y^2+y^4)

Now, putting x=a^2\and\:y=\sqrt{a^2-1} we get,

\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[(a^2)^4+6(a^2)^2(\sqrt{a^2-1})^2+(\sqrt{a^2-1})^4]

\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2[a^8+6a^4(a^2-1)+(a^2-1)^2]

\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-12a^4+2a^4-4a^2+2

\left(a^2 + \sqrt{a^2 -1} \right )^4 + \left(a^2 - \sqrt{a^2 -1} \right )^4=2a^8+12a^6-10a^4-4a^2+2

 

Posted by

Pankaj Sanodiya

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