Question

Factorise x^4-81


With Explanation
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Answer 1

ANSWER

${x}^{4} - 81 \\ \\ = >( x {}^{2} ) {}^{2} - {(9)}^{2} \\ here \: using \: identies \\ \\ = > {x}^{2} - {y}^{2} = (x + y)(x - y) \\ so \\ \\ = > x = {x}^{2} and \: y = 9 \\ \\ = > ( {x}^{2} + 9)( {x}^{2} - 9)$

Answer 2

Answer:

(x-3)(x+3)(x^2+9)

Explanation:-

a^2-b^2=(a-b)(a+b)..................(1)

here(x^2)^2=x^4 and b=(9)^2=81

=>a=x^2 and b=9

substituting into (1)

=>x^4-81=(x-9)(x+9)......................(2)

Now the factor

x^2-9 is also a difference of squares

=>x^2-9=(x-3)(x+3)

substituting into (2) to complete the factorising

x^4-81=(x-3)(x+3)(x^2+9).

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