Let f(x)=x? (cubic) +x- 1

f(0)=- 1 & lt； 0

f( 1)= 1+ 1- 1 = 1 & gt； 0

f'(x)=3x？ (square)+1> 0

Therefore, f(x) increases monotonically at (0, 1).

So there is only one real root in (0, 1).

Complete the certificate.

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