Answer
a. See proof
b. $X^3+17X-\frac{34}{3}$
Work Step by Step
a. Replace x with $X-a/3$ in the equation, we have
$x^3+ax^2+bx+c=(X-\frac{a}{3})^3+a(X-\frac{a}{3})^2+b(X-\frac{a}{3})+c
=X^3-3\frac{a}{3}X^2+3\frac{a^2}{9}X-\frac{a^3}{27}+aX^2-2\frac{a}{3}X+a\frac{a^2}{9}+bX-\frac{ab}{3}+c
=X^3+(\frac{a^2}{3}-\frac{2a}{3}+b)X+(\frac{2a^3}{27}-\frac{ab}{3}+c)$ which shows that the $x^2$ term disappeared.
b. Following the results from (a) we have $a=6,b=9,c=4$ and
$x^3+6x^2+9x+4=X^3+(6^2/3-2\times6/3+9)X+(2\times6^2/27-6\times9/3+4)
=X^3+17X-\frac{34}{3}$