1. Since $\alpha,\beta,\gamma$ are three consecutive terms of a non-constant G.P., we have $\beta^2=\alpha\gamma$ and the common ratio is not $1$.

2. Let the common root of $\alpha x^2+2\beta x+\gamma=0$ and $x^2+x-1=0$ be $r$.

Then $r$ satisfies $r^2+r-1=0 \implies r^2=1-r.$

3. Substitute this into the first equation: $\alpha r^2+2\beta r+\gamma=0$ $\alpha(1-r)+2\beta r+\gamma=0$ $\alpha+\gamma+r(2\beta-\alpha)=0.$

So, $r=-\frac{\alpha+\gamma}{2\beta-\alpha}.$

But also $r$ is a root of $x^2+x-1=0$, so eliminating $r$ is easier by using the fact that any common root implies the two quadratics have a common factor.

Since $x^2+x-1=0$ is irreducible over general coefficients and has roots $r_1,r_2$, let one of them be common. Write $\alpha x^2+2\beta x+\gamma = k(x-r)(x-s),$ with $r$ a root of $x^2+x-1=0$.

A cleaner way is to use the relation for roots of $x^2+x-1=0$: If $r$ is a root, then the other root is $-1-r$ and $r+\frac1r=-1.$ Also from the first equation, $\alpha r^2+2\beta r+\gamma=0.$ Dividing by $r$ (since $r\neq 0$), $\alpha r+2\beta+\frac{\gamma}{r}=0.$

Using G.P., let $\beta=\alpha q,\qquad \gamma=\alpha q^2,$ with $q\neq 1$. Then the first equation becomes $\alpha(x^2+2qx+q^2)=0$ $\alpha(x+q)^2=0.$ So its only root is $x=-q.$

4. Since this has a common root with $x^2+x-1=0$, we must have $(-q)^2+(-q)-1=0$ $q^2-q-1=0.$

5. Now compute $\alpha(\beta+\gamma)=\alpha(\alpha q+\alpha q^2)=\alpha^2(q+q^2)=\alpha^2 q(1+q).$

From $q^2-q-1=0$, $q^2=q+1 \implies 1+q=q^2.$ Hence $\alpha(\beta+\gamma)=\alpha^2 q\cdot q^2=\alpha^2 q^3.$

Now, $\beta\gamma=(\alpha q)(\alpha q^2)=\alpha^2 q^3.$ Therefore, $\boxed{\alpha(\beta+\gamma)=\beta\gamma}.$

6. Checking options:

• A: $\alpha\gamma=\alpha^2q^2$ — not equal in general
• B: $0$ — not true in general
• C: $\beta\gamma$ — correct
• D: $\alpha\beta=\alpha^2q$ — not equal in general

So the correct option is C.