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(B) The given polynomial equation is $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. This expression follows the binomial expansion of $(x-1)^6$. Thus,the equati

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$\frac{2}{3}$

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(B) The given polynomial equation is $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. This expression follows the binomial expansion of $(x-1)^6$. Thus,the equation can be written as $(x-1)^6=0$. This implies that all roots of the equation are equal to $1$. Therefore,the largest real root $a = 1$ and the smallest real root $b = 1$. Substituting these values into the expression,we get $\frac{a^2+b^2}{a+b+1} = \frac{1^2+1^2}{1+1+1} = \frac{1+1}{3} = \frac{2}{3}$.

Let $a$ be the largest real root and $b$ be the smallest real root of the polynomial equation $x^6-6x^5+15x^4-20x^3+15x^2-6x+1=0$. Then $\frac{a^2+b^2}{a+b+1}$ is

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