Adding cubes of the roots of solution of a cubic equation

I was solving a question paper for pg entrance exam in an university?It is an mcq question So i came to this question .I don't know which method to use The question is....

If  $\alpha ,\beta,\gamma,$ are the roots of the equation $$15x^3+7x-11=0$$ then the value of $\alpha ^3+\beta^3+\gamma^3$ is (the options are)

1. $\frac{3}{5}$
2. $\frac{7}{5}$
3. $\frac{9}{5}$
4. $\frac{11}{5}$

I did like this $$15x^3+7x-11=0$$ $$x^3=\frac{11-7x}{15}$$

And substituted $\alpha ,\beta,\gamma,$ and added the three equations but I am not getting a real number.the answer is a real number

So you get $$\alpha^3+\beta^3+\gamma^3=\frac{11-7\alpha}{15} +\frac{11-7\beta}{15}+\frac{11-7\gamma}{15} =\frac{11}5-\frac7{15}(\alpha+\beta+\gamma).$$ The value of $\alpha+\beta+\gamma$ should be immediately accessible from the cubic equation...
• Do you know how to write down the value of $\alpha+\beta+\gamma$ given a cubic equation? – Lord Shark the Unknown Jun 3 '17 at 3:49
• @NBN If you have a cubic equation $x^3+px^2+qx+r=0$ with roots $\alpha$, $\beta$, $\gamma$, then we have the factorisation $x^3+px^2+qx+r=(x-\alpha)(x-\beta)(x-\gamma)$. Comparing $x^2$ coefficients: $p=-\alpha-\beta-\gamma$. – Lord Shark the Unknown Jun 3 '17 at 3:55
For all $x$ we have $15x^3+7x-15=15(x-a)(x-b)(x-c).$
Equating coefficients, the coefficient of $x^2$ on the RHS above, which is $-15(a+b+c),$ must be $0.$ Therefore $a+b+c=0$, so $$a^3+b^3+c^3=(1/15)((-7a+11)+(-7b+11)+(-7c+11))=33/15=11/5.$$