Let $t$ be TNYWR.

The expression $(t x+3)^{3}$ can be re-written in the form $a x^{3}+b x^{2}+c x+d$ for some positive integers $a, b, c, d$. Determine the value of the largest of $a, b, c$, and $d$.
Expanding,

$$
(t x+3)^{3}=(t x+3)(t x+3)^{2}=(t x+3)\left(t^{2} x^{2}+6 t x+9\right)=t^{3} x^{3}+9 t^{2} x^{2}+27 t x+27
$$

In terms of $t$, the coefficients are $a=t^{3}, b=9 t^{2}, c=27 t$, and $d=27$.

Since the answer to (b) is 6 , then $t=6$, and so the coefficients are $a=6^{3}=216$, $b=9(36)=324, c=27(6)=162$, and $d=27$.

Therefore, the largest of the values of $a, b, c, d$ is 324 .

ANswer: 9, 6,324