Question

${2}^{1 \div 3} - {2}^{ - 1 \div 3} = x \\ find \: the \: value \: of \: 2 {x}^{3} + 6x$

Answer 1

ANSWER--


x = 2^(1/3) - 2^(-1/3) 
= 2^(1/3) - 1/2^(1/3) 
= (2^(2/3) -1)/2^(1/3) 

therefore, 2x^3 + 6x 
= 2 [2x^(2/3) -1]^3 + 6(2^(2/3) -1)/2^(1/3) 
= 2^2 -1 -3*2^(2/3)(2^(2/3) -1) + 6*2^(2/3)(2^(2/3) -1) 
= 3 - 3 * 2 ^(4/3) + 3*2^(2/3) + 3*2^(2/3)(2^(2/3)-1) 
= 3 - 3*2*2^(1/3) + 3*2^(2/3) + 3*2*2^(1/3) - 3*2^(2/3) 
= 3