Question

X=2⅓-2-⅓,prove 2x³+6x-3=0

Answer 1

$x = {2}^{ \frac{1}{3} } - 2 ^{ \frac{ - 1}{3} } \\ cubing \: \: on \: both \: sides \\ {x}^{3} = ({2}^{ \frac{1}{3} } - 2 ^{ \frac{ - 1}{3} }) ^{3} \\ {x}^{3} = ( {2}^{ \frac{1}{3} } )^{3} - ( {2}^{ \frac{ - 1}{3} } )^{3} - 3(2 ^{ \frac{1}{3} })(2 ^{ \frac{ - 1}{3} })( {2}^{ \frac{1}{3} } - 2 ^{ \frac{ - 1}{3} } )\\ {x}^{3} = 2 - {2}^{ - 1} - 3(x) \\ {x}^{3} = 2 - \frac{1}{2} - 3x \\ {2x}^{3} = 4 - 1 - 6x \\ {2x}^{3} = 3 - 6x \\ {2x}^{3} + 6x - 3 = 0 \\ hence \: proved$