solve this pls its kinda hard

By Vieta's formulas, the sum of the roots of f(x) is −5, so the sum of the roots of x3+3x−3=0 is −(r1​+r2​+⋯+r7​)=5.

By the Factor Theorem, x−5 is a factor of x3+3x−3. We can use polynomial division (or inspection) to find that x3+3x−3=(x−5)(x2+8x+3). Then

$$\begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (r_1 - 5)(r_1^2 + 8r_1 + 3) (r_2 - 5)(r_2^2 + 8r_2 + 3) \dotsm (r_7 - 5)(r_7^2 + 8r_7 + 3) \ &= (r_1 - 5)(r_2 - 5) \dotsm (r_7 - 5) (r_1^2 + r_2^2 + \dots + r_7^2 + 8(r_1 + r_2 + \dots + r_7) + 21). \end{align*}$$

By Vieta's formulas again, r12​+r22​+⋯+r72​=49−2⋅4+2⋅5−9+1=43, and r1​+r2​+⋯+r7​=−5.

Hence, $$\begin{align*} g(r_1) g(r_2) \dotsm g(r_7) &= (-5)(-5) \dotsm (-5) (43 + 8 \cdot -5 + 21) \ &= \boxed{125a + 3125}. \end{align*}$$