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umka21 [38]
3 years ago
5

Factor completly x^8- 1/81. please help:)

Mathematics
2 answers:
Leno4ka [110]3 years ago
5 0

Answer:

{x}^{8}  -  \frac{1}{81}  = ( {x}^{4}  +  \frac{1}{9} )(  {x}^{2}  +  \frac{1}{3} )(x +  \frac{1}{ \sqrt{3} } )(x -  \frac{1}{ \sqrt{3} } )

Step-by-step explanation:

see the picture attached for further explanation.

Lena [83]3 years ago
4 0

Answer:

\large\boxed{\left(x^4+\dfrac{1}{9}\right)\left(x^2+\dfrac{1}{3}\right)\left(x+\dfrac{\sqrt3}{3}\right)\left(x-\dfrac{\sqrt3}{3}\right)}

Step-by-step explanation:

Use\\\\a^2-b^2=(a-b)(a+b)\\\\(a^n)^m=a^{nm}\\\\x^8-\dfrac{1}{81}=x^{4\cdot2}-\dfrac{1}{9^2}=(x^4)^2-\left(\dfrac{1}{9}\right)^2=\left(x^4+\dfrac{1}{9}\right)\underbrace{\left(x^4-\dfrac{1}{9}\right)}_{(*)}\\\\(*)=x^{2\cdot2}-\dfrac{1}{3^2}=(x^2)^2-\left(\dfrac{1}{3}\right)^2=\left(x^2+\dfrac{1}{3}\right)\underbrace{\left(x^2-\dfrac{1}{3}\right)}_{(**)}\\\\(**)=x^2-\dfrac{3}{9}=x^2-\dfrac{(\sqrt3)^2}{3^2}=x^2-\left(\dfrac{\sqrt3}{3}\right)^2=\left(x+\dfrac{\sqrt3}{3}\right)\left(x-\dfrac{\sqrt3}{3}\right)

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