anonymous
  • anonymous
Solve x3 = 64 over 27. ±8 over 3 8 over 3 ±4 over 3 4 over 3
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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anonymous
  • anonymous
@dan815 @Nnesha
Nnesha
  • Nnesha
take cube root both sides to cancel out the cube
anonymous
  • anonymous
how

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Nnesha
  • Nnesha
here is an example \[\huge\rm \sqrt[3]{m^3} = \sqrt[3]{27}\]
anonymous
  • anonymous
O.o
anonymous
  • anonymous
im confuzzled
Nnesha
  • Nnesha
you can convert root to exponent cube root of m^3 is same as \[\huge\rm m^{3 \times \frac{ 1 }{ 3 }}\] then you can cancel out the 3's \[\huge\rm m^{\cancel{3} \times \frac{ 1 }{ \cancel{3 }}}=m\] that's how you can take cube root both sides to cancel out cube
Nnesha
  • Nnesha
take cube root both sides !
anonymous
  • anonymous
*cries*
anonymous
  • anonymous
idk how
Nnesha
  • Nnesha
\[\huge\rm \sqrt[3]{x^3}=\sqrt[3]{\frac{ 64 }{ 27 }}\] that's how we should take cube root cube= 3
Nnesha
  • Nnesha
\[\huge\rm \sqrt[3]{\frac{ a }{ b }}=\frac{ \sqrt[3]{a} }{ \sqrt[3]{b} }\]
Nnesha
  • Nnesha
take cube root of the numerator and take cube root of the denominator !

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