Find the sum of the cubes of the roots of the equation $3y^2- 14y + 8 = 0$:
- $\cfrac{1728}{27}$
- $\cfrac{1736}{27}$
- $\cfrac{1730}{27}$
- $\cfrac{1732}{27}$
Reveal answer
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B
Correct answer
Explanation
For 3y^2 - 14y + 8 = 0, sum of roots alpha + beta = 14/3 and product alpha * beta = 8/3. The sum of cubes is (alpha + beta)^3 - 3 * alpha * beta * (alpha + beta) = (14/3)^3 - 3 * (8/3) * (14/3) = 2744/27 - 336/9 = 2744/27 - 1008/27 = 1736/27.
AI explanation
Using Vieta's formulas for 3y^2 - 14y + 8 = 0, the sum of the roots (y1 + y2) is 14/3 and the product (y1 * y2) is 8/3. We use the identity for the sum of cubes, which states y1^3 + y2^3 = (y1 + y2)^3 - 3*y1*y2*(y1 + y2). Substituting the known values gives (14/3)^3 - 3*(8/3)*(14/3), which simplifies to 2744/27 - 1008/27. The final result is 1736/27.