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Zinaida [17]
3 years ago
9

Factor completely.

c{1}{81}" alt="x^{8}-\frac{1}{81}" align="absmiddle" class="latex-formula">
Mathematics
1 answer:
Eduardwww [97]3 years ago
6 0

We have 3⁴ = 81, so we can factorize this as a difference of squares twice:

x^8 - \dfrac1{81} = \left(x^2\right)^4 - \left(\dfrac13\right)^4 \\\\ x^8 - \dfrac1{81} = \left(\left(x^2\right)^2 - \left(\dfrac13\right)^2\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(\left(x^2\right)^2 + \left(\dfrac13\right)^2\right) \\\\ x^8 - \dfrac1{81} = \left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)

Depending on the precise definition of "completely" in this context, you can go a bit further and factorize x^2-\frac13 as yet another difference of squares:

x^2 - \dfrac13 = x^2 - \left(\dfrac1{\sqrt3}\right)^2 = \left(x-\dfrac1{\sqrt3}\right)\left(x+\dfrac1{\sqrt3}\right)

And if you're working over the field of complex numbers, you can go even further. For instance,

x^4 + \dfrac19 = \left(x^2\right)^2 - \left(i\dfrac13\right)^2 = \left(x^2 - i\dfrac13\right) \left(x^2 + i\dfrac13\right)

But I think you'd be fine stopping at the first result,

x^8 - \dfrac1{81} = \boxed{\left(x^2 - \dfrac13\right) \left(x^2 + \dfrac13\right) \left(x^4 + \dfrac19\right)}

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