Find the value of $$(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$$

Asked by Pragya Singh | 1 year ago |  111

Solution :-

Firstly, the expression is simplified by using Binomial Theorem

$$(x+y)^4+(x-y)^4$$

This can be done as

$$(x+y)^4=^4C_0x^4+\;^4C_1x^3y+\;$$

$$^4C_2x^2y^2+\;^4C_3xy^3+\;^4C_4y^4$$

$$x^4+4x^3y+6x^2y^2+4xy^3+y^4$$

$$(x-y)^4=\;^4C_0x^4-\;^4C_1x^3y+\;$$

$$^4C_2x^2y^2-\;^4C_3xy^3+\;^4C_4y^4$$

$$x^4-4x^3y+6x^2y^2-4xy^3+y^4$$

Putting $$x=a^2$$ and $$y=\sqrt{a^2-1}$$, we obtain

$$(a^2+\sqrt{a^2-1})^4+(a^2-\sqrt{a^2-1})^4$$

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

$$2[a^8+6a^4(a^2-1)+(a^2-1)^2]$$

$$2[a^8+6a^4-6a^4+a^4-2a^2+1]$$

$$= 2[a^8+6a^6-5a^4-2a^2+1]$$

$$2a^8+12a^6-10a^4-4a^2+2$$

Answered by Abhisek | 1 year ago

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