Let the four consecutive terms of the G.P. be $\frac{a}{r},\ a,\ ar,\ ar^2$ where $a>0$ and $r>0$ since all terms are positive.

Use the product condition Their product is $\frac{a}{r}\cdot a\cdot ar\cdot ar^2 = a^4r^2$ Given product $=1296$, so $a^4r^2=1296 \qquad (1)$

Use the sum condition Their sum is $\frac{a}{r}+a+ar+ar^2=126$ Factor $a$: $a\left(\frac1r+1+r+r^2\right)=126 \qquad (2)$

Simplify using the product equation From (1), $a^4r^2=1296=6^4$ Since all quantities are positive, $ar^{1/2}=6$ but a cleaner approach is to write the four terms symmetrically as $\frac{b}{r^3},\ \frac{b}{r},\ br,\ br^3$ Then their product is $\frac{b}{r^3}\cdot \frac{b}{r}\cdot br\cdot br^3=b^4=1296$ Hence $b=6$

So the terms are $\frac{6}{r^3},\ \frac{6}{r},\ 6r,\ 6r^3$

Apply the sum condition $\frac{6}{r^3}+\frac{6}{r}+6r+6r^3=126$ Divide by $6$: $r^3+r+\frac1r+\frac1{r^3}=21$

Set $x=r+\frac1r$ Then $r^3+\frac1{r^3}=x^3-3x$ So the equation becomes $x^3-3x+x=21$ $x^3-2x-21=0$

Solve the cubic Try integer roots. For $x=3$, $27-6-21=0$ so $(x-3)(x^2+3x+7)=0$ The quadratic has discriminant $9-28=-19<0$ so the only real solution is $x=3$

Thus, $r+\frac1r=3$

Find possible common ratios $r+\frac1r=3 \implies r^2-3r+1=0$ Hence $r=\frac{3\pm\sqrt5}{2}$

Both are positive, so both are valid common ratios.

Sum of all such common ratios $\frac{3+\sqrt5}{2}+\frac{3-\sqrt5}{2}=3$

Compare with options The required sum is $3$ So the correct option is C.

Comparison with stored answer Stored correct answer is A: 7, but our derived answer is C: 3. Hence the stored answer appears to be incorrect.