Consider the expression $$(xxx)_{b}=x^3$$, where b is the base, and x is any digit of base b. Find the value of b:
$$(xxx)_{b}=x^3$$
=> $$xb^2+xb+x = x^3$$
=> $$b^2+b+1=x^2$$
On substituting b=1,and b=2, we get $$x^2$$ as $$3$$, and $$7$$. Since $$3$$ and $$7$$ are not perfect squares, we can infer that no number satisfies the given condition. Therefore, option E is the right answer.
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