The square root of frac{(0.75)^{3}}{1-0.75}+[ | vedclass

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Question

(B) Let $a = 0.75$.The expression is $\frac{a^3}{1-a} + (a^2 + a + 1)$.We know that $1 - a^3 = (1 - a)(1 + a + a^2)$,so $\frac{1 - a^3}{1 - a} = 1 + a

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) Let $a = 0.75$.The expression is $\frac{a^3}{1-a} + (a^2 + a + 1)$.We know that $1 - a^3 = (1 - a)(1 + a + a^2)$,so $\frac{1 - a^3}{1 - a} = 1 + a + a^2$.Substituting this into the expression:Expression $= \frac{a^3}{1-a} + \frac{1-a^3}{1-a} = \frac{a^3 + 1 - a^3}{1-a} = \frac{1}{1-a}$.Substituting $a = 0.75$ back:Expression $= \frac{1}{1 - 0.75} = \frac{1}{0.25} = 4$.The square root of the expression is $\sqrt{4} = 2$.

The square root of $\frac{(0.75)^{3}}{1-0.75}+[0.75+(0.75)^{2}+1]$ is

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