Consider the expression $$\frac{(a^2+a+1)(b^2+b+1)(c^2+c+1)(d^2+d+1)(e^2+e+1)}{abcde}$$, where a,b,c,d and e are positive numbers. The minimum value of the expression is

The given expression can be written as $$\frac{a^2+a+1}{a}*\frac{b^2+b+1}{b}*\frac{c^2+c+1}{c}*\frac{d^2+d+1}{d}*\frac{e^2+e+1}{e}$$.
$$\frac{a^2+a+1}{a}=a+\frac{1}{a}+1$$
We know that for positive values, AM $$\geq$$ GM. 
$$\frac{1+\frac{1}{a}}{2} \geq \sqrt{a*\frac{1}{a}}$$
$$a+\frac{1}{a} \geq 2$$
The least value that $$a+\frac{1}{a}$$ can take is $$2$$.
Therefore, the least value that the term $$a+\frac{1}{a}+1$$ can take is $$3$$.
Similarly, the least value that the other terms can take is also $$3$$.
=> The least value of the given expression = $$3*3*3*3*3$$ = $$243$$.
Therefore, option E is the right answer.