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Solution
$$\frac{(1-d^3)}{(1-d)}$$ = $$1+d^2+d$$ (where $$d \neq 1$$)
Let's say $$f(d)=1+d^2+d$$
Now $$f(d)$$ will always be greater than 0 and have its minimum value at d = -0.5. The value is $$\frac{3}{4}$$.
For $$d<-1$$ ; $$f(d) >1$$
$$-1<d<0$$ ; $$\frac{3}{4} <f(d)<1$$
$$0<d<1$$ ; $$1<f(d)<3$$
$$d>1$$ ; $$f(d)>3$$
So, for d > 1, f(d) > 3. Option b) is the correct answer.
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