Question

x=1+2^1/3+2^2/3 show that x^3-3x^2-3x-1=0

Answer 1

$x = 1 + \sqrt[3]{2} + \ \sqrt[3]{ {2}^{2} } \\ x - 1 = sqrt[3]{2} + \ \sqrt[3]{ {2}^{2} } \\ ({x - 1})^{3} = ( {sqrt[3]{2} + \ \sqrt[3]{ {2}^{2} } })^{3} \\ {x}^{3 } - {3x}^{2} + 3x - 1 = 6 + 3 \sqrt[3]{8} ( {sqrt[3]{2} + \ \sqrt[3]{ {2}^{2} } }) \\ {x}^{3 } - {3x}^{2} + 3x - 1 = 6 + 6x - 6 \\ {x}^{3 } - {3x}^{2} - 3x - 1 = 0$

Answer 2

Answer:

this is answer and hope this helps you thanks