Question

(a-1/a)(a+1/a)(a²+1/a²)(a power 4+1/a power 4)

Answer 1

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

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

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

$( { {a}^{4} )}^{2} - {( { \frac{1}{a} }^{4}) }^{2} \\ \\ so \: final \: answer \: is \: \\ \\ {a}^{8} - {( \frac{1}{a}) }^{8}$

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Answer 2

Answer: $(a^8-\frac{1}{a^8})$

Step-by-step explanation:

Given expression: $(a-\frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})$

Since, we know that  $(a+b)(a-b)=(a^2-b^2)$

Therefore, we get

$(a-\frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})\\\\=(a^2-\frac{1}{a^2})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})..........[Since,\ (a-\frac{1}{a})(a+\frac{1}{a})=(a^2-\frac{1}{a^2})]\\\\=(a^4-\frac{1}{a^4})(a^4+\frac{1}{a^4})..........[Since,\ (a^2-\frac{1}{a^2})(a^2+\frac{1}{a^2})=(a^4-\frac{1}{a^4})]\\\\=((a^4)^2-\frac{1}{(a^4)^2})\\\\=(a^8-\frac{1}{a^8})$

Hence, $(a-\frac{1}{a})(a+\frac{1}{a})(a^2+\frac{1}{a^2})(a^4+\frac{1}{a^4})=(a^8-\frac{1}{a^8})$