x-3-1-2-2x-1-1-3-

Question Number 140669 by john_santu last updated on 11/May/21

x^{3} + 1 = 2 \sqrt[3]{2 x - 1}

Answered by bemath last updated on 11/May/21

\frac{x^{3} + 1}{2} = \sqrt[3]{2 x - 1}

set g (x) = \sqrt[3]{2 x - 1} \Rightarrow \frac{x^{3} + 1}{2} = g^{- 1} (x)

\Rightarrow g (g^{- 1} (x)) = x

\Rightarrow \frac{x^{3} + 1}{2} = x; x^{3} - 2 x + 1 = 0

\Rightarrow (x - 1) (x^{2} + x - 1) = 0

\Rightarrow x = 1 \land \frac{- 1 \pm \sqrt{5}}{2}

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