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x^6 = 27x^3
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\[ x^6-27x^3 = 0 \\ x^3(x^3-27) = 0 \]
Whats the solution though ? I got the x3(x3−27)=0 idk what to do after that
x=?
Two equations: \[ x^3=0\\ x^3-27=0 \quad \quad x^3=27 \]So there are 2 unique solutions.
Do I not get rid of the 3 in the x
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Take the cube root \(\sqrt[3]{\quad}\)
so 0 and 3
Yes.
THanks !
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