The square root of left[frac{(3.25)^{3}}{3.25 | vedclass

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(B) Let $a = 3.25$.The given expression is $E = \left[\frac{a^3}{a-1} - (a^2 + a + 1)\right]$.We know that $a^3 - 1 = (a - 1)(a^2 + a + 1)$,which impl

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$\frac{2}{3}$

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(B) Let $a = 3.25$.The given expression is $E = \left[\frac{a^3}{a-1} - (a^2 + a + 1)\right]$.We know that $a^3 - 1 = (a - 1)(a^2 + a + 1)$,which implies $\frac{a^3 - 1}{a - 1} = a^2 + a + 1$.Substituting this into the expression $E$:$E = \frac{a^3}{a-1} - \frac{a^3 - 1}{a-1} = \frac{a^3 - (a^3 - 1)}{a-1} = \frac{1}{a-1}$.Now,substitute $a = 3.25$:$E = \frac{1}{3.25 - 1} = \frac{1}{2.25} = \frac{1}{9/4} = \frac{4}{9}$.The square root of the expression is $\sqrt{E} = \sqrt{\frac{4}{9}} = \frac{2}{3}$.

The square root of $\left[\frac{(3.25)^{3}}{3.25-1}-\left(3.25+(3.25)^{2}+1\right)\right]$ is equal to

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