The value of a (a ge 3) for which the sum of | vedclass

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(A) Let the roots of the quadratic equation $x^2 - (a - 2)x + (a - 3) = 0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \bet

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$3$

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(A) Let the roots of the quadratic equation $x^2 - (a - 2)x + (a - 3) = 0$ be $\alpha$ and $\beta$.From the properties of roots,we have $\alpha + \beta = a - 2$ and $\alpha \beta = a - 3$.The sum of the cubes of the roots is given by $S = \alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3\alpha \beta (\alpha + \beta)$.Substituting the values,$S = (a - 2)^3 - 3(a - 3)(a - 2)$.Expanding this expression: $S = (a^3 - 6a^2 + 12a - 8) - 3(a^2 - 5a + 6) = a^3 - 6a^2 + 12a - 8 - 3a^2 + 15a - 18 = a^3 - 9a^2 + 27a - 26$.We can rewrite this as $S = (a - 3)^3 + 1$.Since $a \ge 3$,the term $(a - 3)^3$ is non-negative and its minimum value is $0$ when $a = 3$.Thus,the sum of the cubes of the roots assumes the least value when $a = 3$.

The value of $a$ $(a \ge 3)$ for which the sum of the cubes of the roots of $x^2 - (a - 2)x + (a - 3) = 0$ assumes the least value is:

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