Using sum/product of roots of a cubic equation to solve given expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ If $a, b, c$ are the roots of the equation $7x^3- 25x +42 =0$, then the value of the expression $(a+b)^3+ (b+c)^3 + (c+a)^3$ is?
I tried to solve this but wasn't able to simplify the term to be able to apply sum and product of roots of equations. Could someone help me through simplification of the expression to make this solvable?
Since $a+b+c = \frac{-25}{7}$, $ab + bc + ca = 0$ and $abc = -42$, if i factorize the expression i get
$(2[(a+b+c)^3 - 3a^2(b+c) - 3 c^2 (a+b) -3b^2(a+c) -6abc]) - 3(a+b)^2(b+2c+a) - 3(c+a)^2(c+2b+a) - 3(b+c)^2(c+2a+b))$
This is what i get, after my best attempts at simplification. Since this is an mcq type question, i believe this does not require such mammoth efforts. Could someone provide me a hint and possibly an easier way of simplification/application of the concept?

Using Vieta's formula $\displaystyle a+b+c=0,(-1)^3abc=42$
$$\implies \sum (a+b)^3=-\sum c^3$$
Method $\#1:$
As $a,b,c$ individually satisfy the given equation, we can write $$7a^3=25a-42\text{ etc.}$$
Again using Newton's 
Power Sum formula $$7\sum a^3=25\sum a-3\cdot42=\cdots$$
Method $\#2:$
Like If $a,b,c \in R$ are distinct, then $-a^3-b^3-c^3+3abc \neq 0$.,
$$a^3+b^3+c^3=3abc$$
