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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