
Find the sum of the cubes of the roots of the equation $3{y^2} - 14y + 8 = 0$.
$
A.\dfrac{{1728}}{{27}} \\
B.\dfrac{{1736}}{{27}} \\
C.\dfrac{{1730}}{{27}} \\
D.\dfrac{{1732}}{{27}} \\
$
Answer
564.9k+ views
Hint: The cube of a number is defined as the multiplication of the number by itself three times. In other words, when a number is multiplied by itself three times, it results in the cube of a number. The roots of a quadratic equation can be found by several methods like the try and error method, splitting the middle term method, etc.
Here, we are determining the roots of the equation by splitting the middle terms as the coefficients are so large. In this method, the middle term i.e, the coefficient of x is splitted in two terms such that the sum of the terms results in the coefficient of x while the product of the terms is the product of the coefficients of ${x^2}$ and ${x^0}$.
Complete step by step solution:
To determine the cubes of the roots of the equation $3{y^2} - 14y + 8 = 0$, we need to first determine the roots of the equation $3{y^2} - 14y + 8 = 0$ by splitting the middle term method.
$ 3{y^2} - 14y + 8 = 0 \\
3{y^2} - (12 + 2)y + 8 = 0 \\
3{y^2} - 12y - 2y + 8 = 0 \\
3y(y - 4) - 2(y - 4) = 0 \\
(3y - 2)(y - 4) = 0 \\
y = \dfrac{2}{3};4 \\ $
Now, we know about the roots of the equation $3{y^2} - 14y + 8 = 0$ which is $\dfrac{2}{3}{\text{ and, 4}}$. So, getting the cubes of the roots as:
$ {\left( {\dfrac{2}{3}} \right)^3} = \dfrac{8}{{27}} \\
{4^3} = 64 \\ $
Now, the sum of the cube of the roots is calculated as:
$ S = \dfrac{8}{{27}} + 64 \\
= \dfrac{{8 + (64 \times 27)}}{{27}} \\
= \dfrac{{8 + 1728}}{{27}} \\
= \dfrac{{1736}}{{27}} \\ $
Hence, the sum of the cubes of the roots of the equation $3{y^2} - 14y + 8 = 0$ is $\dfrac{{1736}}{{27}}$.
Option B is correct.
Note: Don’t confuse the statement of the question about “the cubes of the sum” with the “sum of the cubes”, these have two completely different meanings. Candidates often make mistakes while reading the question so read the question carefully.
Here, we are determining the roots of the equation by splitting the middle terms as the coefficients are so large. In this method, the middle term i.e, the coefficient of x is splitted in two terms such that the sum of the terms results in the coefficient of x while the product of the terms is the product of the coefficients of ${x^2}$ and ${x^0}$.
Complete step by step solution:
To determine the cubes of the roots of the equation $3{y^2} - 14y + 8 = 0$, we need to first determine the roots of the equation $3{y^2} - 14y + 8 = 0$ by splitting the middle term method.
$ 3{y^2} - 14y + 8 = 0 \\
3{y^2} - (12 + 2)y + 8 = 0 \\
3{y^2} - 12y - 2y + 8 = 0 \\
3y(y - 4) - 2(y - 4) = 0 \\
(3y - 2)(y - 4) = 0 \\
y = \dfrac{2}{3};4 \\ $
Now, we know about the roots of the equation $3{y^2} - 14y + 8 = 0$ which is $\dfrac{2}{3}{\text{ and, 4}}$. So, getting the cubes of the roots as:
$ {\left( {\dfrac{2}{3}} \right)^3} = \dfrac{8}{{27}} \\
{4^3} = 64 \\ $
Now, the sum of the cube of the roots is calculated as:
$ S = \dfrac{8}{{27}} + 64 \\
= \dfrac{{8 + (64 \times 27)}}{{27}} \\
= \dfrac{{8 + 1728}}{{27}} \\
= \dfrac{{1736}}{{27}} \\ $
Hence, the sum of the cubes of the roots of the equation $3{y^2} - 14y + 8 = 0$ is $\dfrac{{1736}}{{27}}$.
Option B is correct.
Note: Don’t confuse the statement of the question about “the cubes of the sum” with the “sum of the cubes”, these have two completely different meanings. Candidates often make mistakes while reading the question so read the question carefully.
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