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

Question Number 124452 by liberty last updated on 03/Dec/20

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

Commented by Dwaipayan Shikari last updated on 03/Dec/20

G o l d e n r a t i o ϕ = \frac{1 + \sqrt{5}}{2} = 1.6180 . .

Answered by mnjuly1970 last updated on 03/Dec/20

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

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

f^{- 1} o f = I_{x} = x

2 x + 1 = x^{3}

x^{3} - 2 x - 1 = 0

x^{3} - x - x - 1 = 0

x (x - 1) (x + 1) - (x + 1) = 0

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

x = - 1 & x = \frac{1 \underline{+} \sqrt{5}}{2} ✓

Commented by MJS_new last updated on 03/Dec/20

something went wrong . x = 1 obviously is no

solution

Commented by mnjuly1970 last updated on 03/Dec/20

t h a n k y o u m r m j s . .

y o u a r e r i g h t . .

i m a d e a m i s t a k e i n t h e

c a l u l a t i o n (d e c o m p o s i t i o n)

Commented by mnjuly1970 last updated on 03/Dec/20