Question

( 1 + a ) ( 1- a) ( 1 + a²) ( 1 + a⁴) ( 1+ a⁸)
Solve plz​

Answer 1

Answer:

16

Step-by-step explanation:

karo aisa duniya karna chahe apke jaisa

Answer 2

Final answer:  $( 1 + a ) ( 1- a) ( 1 + a^{2}) ( 1 + a^{4}) ( 1+ a^{8}) = 1-a^{16}$

Given that: We are given  $( 1 + a ) ( 1- a) ( 1 + a^{2}) ( 1 + a^{4}) ( 1+ a^{8})$

To find: We have to find the solution of,

$( 1 + a ) ( 1- a) ( 1 + a^{2}) ( 1 + a^{4}) ( 1+ a^{8})$

Explanation:

• In order to solve  $( 1 + a ) ( 1- a) ( 1 + a^{2}) ( 1 + a^{4}) ( 1+ a^{8})$ by term by term.

• First we solve (1 + a)(1 - a). Then it multiplied with next term (1 + a²), the result is multiplied with next term (1 + a⁴). Finally the product is multiplied with (1 + a⁸) which is the final value.

• We know that,

   $(x - a)(x + a) = x^{2} - a^{2}$

• By using above formula:

$(1 + a)(1 - a) = 1^{2} - a^{2} = 1 -a^{2} \\\\(1 + a^{2})(1 - a^{2}) = 1^{2} - (a^{2} )^{2} = 1 -a^{4} \\\\(1 + a^{4})(1 - a^{4}) = 1^{2} - (a^{4} )^{2} = 1 - a^{8}\\\\(1 + a^{8})(1 - a^{8}) = 1^{2} - (a^{8} )^{2} = 1 - a^{16}$

• $( 1 + a ) ( 1- a) ( 1 + a^{2}) ( 1 + a^{4}) ( 1+ a^{8}) = 1-a^{16}$

To know more about the concept please go through the links

https://brainly.in/question/24525316

https://brainly.in/question/12746110

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