# Hardness of a variant of the CDH problem

Given $$g$$, a generator of a multiplicative group (over some finite field or elliptic curve), and the group elements $$\left( g^x, g^a, g^b, g^c, g^{x(a+b)}, g^{x(b+c)} \right)$$, is possible to efficiently find the value of $$g^{x(a+b+c)}$$ (without knowledge of the values $$x, a, b, c$$)?

I believe the problem at hand is closely related to the CDH problem (given $$\left (g, g^a, g^b \right)$$, find $$g^{ab}$$). An efficient algorithm of CDH immediately leads to an efficient algorithm for the above problem. So the above problem is at least not harder than CDH. However, I neither found a way to use additional information to arrive at an efficient solution nor was I able show that it is in fact as hard as CDH. So any help is highly appreciated.

Let suppose $$\mathcal{B}$$ knows how to compute $$g^{x(a+b+c)}$$, and I want to solve the cdh challenge $$(g,X,Y)$$, (we will interpret $$X$$ as $$g^x$$ and $$Y$$ as $$g^b$$) we choose scalars $$d,e$$ which correspond to $$(a+b)$$ and $$(b+c)$$ and we compute $$Z=\mathcal{B}(X, g^d\cdot Y^{-1}, Y, g^e\cdot Y^{-1},X^d, X^e )$$.
We return $$\frac{X^{d+e}}{Z}$$.
Proof: $$DLog \left(\frac{X^{d+e}}{Z}\right) = DLog \left(X^{d+e}\right) - DLog \left(Z\right) = x(d+e)- x\left( d-b + b + e-b\right) = xb$$.