Question

Find the sum of the cubes of the roots of the equation $3y^2-14y+8=0$:

• A. $\frac{1728}{27}$
• B. $\frac{1736}{27}$
• C. $\frac{1730}{27}$
• D. $\frac{1732}{27}$

Answer 1

The correct option is B $\frac{1736}{27}$

The given equation is $3y^2-14y+8=0$

Let the roots be $a,b$

Given that the sum of roots is $a+b=\frac{14}{3}$ and product of roots is $ab=\frac{8}{3}$

By squaring first equation , we get $a^2+b^2+2ab=\frac{196}{9}$

$\Rightarrow a^2+b^2=\frac{196}{9}-\frac{16}{3}=\frac{148}{9}$

Sum of cubes of roots is $a^3+b^3=(a+b)(a^2+b^2-ab)$

$\Rightarrow a^3+b^3=\left(\frac{14}{3}\right)(\frac{148}{9}-\frac{8}{3})=\frac{124\times 14}{27}=\frac{1736}{27}$

Therefore the correct option is $B$