Multiple choice

If a and b are the roots of the quadratic equation $\displaystyle { 2x }^{ 2 }-6x+3=0$, find the value of $\displaystyle { a }^{ 3 }+{ b }^{ 3 }-3ab\left( { a }^{ 2 }+{ b }^{ 2 } \right) -3ab\left( a+b \right) $.

  1. -27

  2. -25

  3. 27

  4. None of these
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A Correct answer
Explanation

For 2x^2 - 6x + 3 = 0, a+b = 3, ab = 1.5. The expression is a^3 + b^3 - 3ab(a^2 + b^2) - 3ab(a+b). Note that a^3 + b^3 = (a+b)(a^2 - ab + b^2) = (a+b)((a+b)^2 - 3ab). a^3 + b^3 = 3 * (9 - 4.5) = 13.5. a^2 + b^2 = (a+b)^2 - 2ab = 9 - 3 = 6. Expression = 13.5 - 3(1.5)(6) - 3(1.5)(3) = 13.5 - 27 - 13.5 = -27.

AI explanation

By Vieta's formulas for 2x^2 - 6x + 3 = 0, the sum of the roots a + b = 6/2 = 3 and the product ab = 3/2. The given expression can be factored and simplified using the sum of cubes formula, recognizing it as a^3 + b^3 - 3ab(a + b) - 3ab(a^2 + b^2). We know that a^3 + b^3 - 3ab(a + b) equals (a + b)^3 - 6ab(a + b), which evaluates to 3^3 - 6(3/2)(3) = 27 - 27 = 0. We then calculate the remaining term, -3ab(a^2 + b^2), by finding a^2 + b^2 = (a + b)^2 - 2ab = 3^2 - 2(3/2) = 6, so the term becomes -3(3/2)(6) = -27. Adding the two parts gives 0 + (-27) = -27.

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