Need help please 27x^6-8 factoring the sum or difference of cubes

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Need help please 27x^6-8 factoring the sum or difference of cubes

Mathematics
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\[27x ^{6}-8\]
keep in mind I'm factoring the sum or deference of cubues
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27x^6-8 factoring the sum or difference of cubes
(3x^2)^3-2^3
how did we get there?
\[27x^6-8=3^3(x^2)^3-2^3=(3x^2)^3-2^3\] Now we use the difference of cubes formula that states \(a^3-b^3=(a-b)(a^2+ab+b^2)\)
Can you please delete the huge space post?
howhow?
got it thanks
and thanks for the help

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