If (a ‒ b), a, and (a + b) are zeroes of the polynomial $2x^3-6x^2+5x-7$, then find the value of a.

Given (a ‒ b), a, and (a + b) are zeroes of the polynomial, $2x^3-6x^2+5x-7$

For any cubic polynomial of the form $p(x)=a{x}^3+b{x}^2+cx+d$, with zeroes as $\alpha , \beta$and $\gamma$, the relation between the zeroes and the coefficients is given as,

$\alpha +\beta +\gamma =\frac{-b}{a}$ -----(1)

$\alpha \beta +\beta \gamma +\gamma \alpha =\frac{c}{a}$ -----(2)

$\alpha \beta \gamma =\frac{-d}{a}$ -----(3)

Comparing with the general form, the coefficients of the polynomial are a = 2, b = –6, c = 5 and d = –7, and its zeros are $\alpha =(a-b), \beta =a$ and $\gamma =(a+b)$. 

Substituting the values in equation (1), we have,

$\left(a-b\right)+a+\left(a+b\right)=3$

$\Rightarrow 3a=3$

$\Rightarrow a=1$

So, the value of a is 1.