Question

If $\sf{If \ x=2^{\frac{1}{3}}-2^{\frac{-1}{3}}}$
Prove that: 2x³+6x=3.​

Answer 1

Chotu here is ur answer

$x = {2}^{ \frac{1}{3} } - {2}^{ \frac{1}{3} } \\ cube \: on \: both \: side \\ {x}^{3} =( { {2}^{ \frac{1}{3} } - {2}^{ \frac{ - 1}{3} } )}^{3} \\ {x}^{3} = { {2}^{ (\frac{1}{3}) } }^{3} - { {2}^{ (\frac{ - 1}{3}) } }^{3} - 3 \times {2}^{ \frac{1}{3} } \times {2}^{ \frac{ - 1}{3} } ( {2}^{ \frac{ 1}{3} } - {2}^{ \frac{1}{3} } ) \\ {x}^{3} = 2 - \frac{1}{2} - 3x \\ {x}^{3 } + 3x = \frac{3}{2} \\ 2 {x}^{3} + 6x = 3 \\ hence \: proved$