Question

factorise: -
x^8+x^4/16+1/256
​

Answer 1

$\underline{\textsf{Given:}}$

$\mathsf{x^8+\dfrac{x^4}{16}+\dfrac{1}{256}}$

$\underline{\textsf{To find:}}$

$\textsf{Factor of}\;\mathsf{x^8+\dfrac{x^4}{16}+\dfrac{1}{256}}$

$\underline{\textsf{Solution:}}$

$\textsf{Consider,}$

$\mathsf{x^8+\dfrac{x^4}{16}+\dfrac{1}{256}}$

$\textsf{This can be written as}$

$\mathsf{=(x^8+2\dfrac{x^4}{16}+\dfrac{1}{256})-\dfrac{x^4}{16}}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{a^2+2ab+b^2=(a+b)^2}}$

$\mathsf{=(x^4+\dfrac{1}{16})^2-\dfrac{x^4}{16}}$

$\mathsf{=(x^4+\dfrac{1}{16})^2-(\dfrac{x^2}{4})^2}$

$\textsf{Using the identity,}$

$\boxed{\mathsf{a^2-b^2=(a-b)(a+b)}}$

$\mathsf{=(x^4+\dfrac{1}{16}-\dfrac{x^2}{4})\,(x^4+\dfrac{1}{16}+\dfrac{x^2}{4})}$

$\underline{\textsf{Answer:}}$

$\mathsf{x^8+\dfrac{x^4}{16}+\dfrac{1}{256}=(x^4+\dfrac{1}{16}-\dfrac{x^2}{4})\,(x^4+\dfrac{1}{16}+\dfrac{x^2}{4})}$

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Answer 2

Answer:

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and 12 thanks also.