Question

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Factorise the Following -
${x}^{8} - 1$
CLUE -
$use \: identities \: = {a}^{2} - {b}^{2} = > (a + b)(a - b) \: and \\ = > {a}^{4} + {b}^{4} = ( {a}^{2} + \sqrt{2} ab + b {}^{2} )( {a}^{2} - \sqrt{2} ab + {b}^{2} )$
REFERENCE -
Class 9 - RS AGGARWAL - Factorisation of Polynomials - EXERCISE 3B - Q. 37

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Answer 1

Step-by-step explanation:

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Answer 2

GIVEN :-

• x⁸ - 1.

TO FIND :-

• The factorise form of x⁸ - 1.

SOLUTION :-

☯ $\\ \underline{\boldsymbol{According\: to \:the\: Question\: \div }}$

$: \implies \displaystyle \sf \: x ^{8} - 1$

$: \implies \displaystyle \sf(x ^{4})^{2} - (1) ^{2}$

$\dag \: \displaystyle \rm \: Identity : a^{2} - b^{2} = (a + b)(a - b)$

$: \implies \displaystyle \sf \: (x ^{4} + 1)(x ^{4} - 1)$

$: \implies \displaystyle \sf \: \bigg \{x ^{4} + 1 \bigg \} \bigg \{ (x ^{2} ) ^{2} + (1) ^{2} \bigg \}$

$\dag \: \displaystyle \rm \: Identity : a^{2} - b^{2} = (a + b)(a - b)$

$: \implies \displaystyle \sf \bigg \{x ^{4} + 1 \bigg \} \bigg \{ (x ^{2} + 1)(x ^{2} - 1) \bigg \}$

$: \implies \displaystyle \sf \bigg \{x ^{4} + 1 \bigg \} \bigg \{ (x ^{2} + 1)(x )^{2} - (1) ^{2} \bigg \}$

$\dag \: \displaystyle \rm \: Identity : a^{2} - b^{2} = (a + b)(a - b)$

$\displaystyle \sf : \implies\bigg \{x ^{4} + 1 \bigg \} \bigg \{ (x ^{2} + 1)(x - 1)(x + 1) \bigg \}$

$: \implies \underline{ \boxed{\displaystyle \sf (x ^{4} + 1)(x ^{2} + 1)(x + 1)(x - 1)}}$

$\therefore \underline{\displaystyle \sf \ Answer \ is \ (x ^{4} + 1)(x ^{2} + 1)(x + 1)(x - 1)}$