#### Answer

$(198x + 99)( x^2 + x + 1)^{98}$

#### Work Step by Step

Original Expression: $F(u) = (1 + x + x^2)^{99}$
$ u = (1 + x + x^2)$
$ F(u) = x^{99}$
Apply the chain rule: $f′(u)\times u′$
$F’(u) = 99(u)^{98} (2x+1)$
$F'(x) = 99(x^2 + x + 1)^{98} (2x+1)$
Simplify:
$(198x + 99)( x^2 + x + 1)^{98}$