Will the modulus function change it all?

case 1 : $3x^2-10x+3=0$ where $|x-3|\ne 0$ .

$\implies x=\dfrac {10 \pm \sqrt {100-36}}{2×3}\implies x=\dfrac {10 \pm 8}{6}\implies x=\frac{18}{6},\frac {2 }{6}=3,\dfrac {1}{3}.$

But $|x-3|=0$ when $x=3$ $\therefore x=\dfrac {1}{3}$ .

case 2 : $|x-3|=1$ where $3x^{2}-10x+3$ is any number.

$|x-3|=1 \implies x-3=\pm 1 \implies x=2,4 \therefore x=2,4.$

case 3 : $|x-3|=-1$ where $3x^{2}-10x+3$ is an even number.

But $|A|\ge 0 \implies |x-3|\ne -1.$

$\therefore x=\dfrac {1}{3},2,4$

$\therefore 3×\dfrac {1}{3}×2×4=\boxed{8}.$

Take log on both sides

At first we need $|x-3|=1$ and is easy to see that two of the solutions are $2$ and $4$ . Now we don't want the expression to become $0^{0}$ so we look for $y$ such that $|y-3|=0$ , so $y=3$ , we evaluate in the exponent $3\cdot 3^{2}-10\cdot 3+3=0$ . So $3$ is solution of the exponent and $x\ne 3$ . As the product of solutions of the exponent is $\frac{3}{3}=1$ and one of the solutions is $3$ , the other is $\frac{1}{3}$ . So $3 \times \frac{2\cdot 4}{3}=2\cdot 4=8$ . An easier way to do it by only mind, I think. Great problem!

The expression can be equal to 1 if and only if $\left| x-3 \right| =1$ or $3{ x }^{ 2 }-10x+3=0$ and $\left| x-3 \right| \neq 0$ .

$\left| x-3 \right| =1$ for $x = 4$ and $x = 2$ .

By factoring $3{ x }^{ 2 }-10x+3=0$ , we find that $3(x-\frac { 1 }{ 3 } )(x-3)=0$ , therefore our possible solutions are $x = \frac{1}{3}$ and $x = 3$ . However, the latter solution does not satisfy the condition $\left| x-3 \right| \neq 0$ , therefore our values of $x$ are 4, 2, and $\frac{1}{3}$ .

$(4 \times 2 \times \frac{1}{3}) \times 3 = \boxed{8}$

Suppose $x \neq 3$ solves the equation. Raise both sides to the power $\frac{1}{x-3}$ to obtain : $|x-3|^{3x-1} = 1.$ Now, if $x=\frac{1}{3}$ , we see that the equation is satisfied. So, this is one solution. Assume, now that $x \neq \frac{1}{3}$ . By a similar process as before, we can raise to the appropriate power to get $|x-3|=1,$ which is found to have solutions : 2 and 4. We can then easily check that these solutions satisfy our equation.

Thus, collecting everything together, we have 3 solutions : $\frac{1}{3}, 2, 4$ . Hence the answer is 8, as required.

We got three values of $x$ say $2,4\frac{1}{3}$ Simply multiply them $3(2×4×\frac{1}{3})$ which is $\boxed{8}$

lol very easy i dont know who rated it as level 4!!!!!!!!. i am of level 2 in algebra THE SOLUTION FOR THIS PROBLEM IS

ROOTS OF THE EQUATION 3X2-1OX+3=0 are 0.33333..... and 3.SO IF THE POWER IS 0 THEN RESULTANT GIVES 1.BUT,0 POWER 0 IS NOT DEFINED SO 3 DOESNT SATISFY. BUT IF THE BASE IS 1 WHAT EVER MAY BE THE POWER RESULTANT WILL BE 1. SO WE HAVE TWO CASES, 1 IS X=4. CASE 2 IS X= 2 SO ANS IS 3 x 2 x 4 x 0.333333....=4 x 2 x 1=8