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...

  • $\begingroup$ Should i do it the same way we found this $\endgroup$ – NBN Jun 3 '17 at 3:47
  • $\begingroup$ Do you know how to write down the value of $\alpha+\beta+\gamma$ given a cubic equation? $\endgroup$ – Lord Shark the Unknown Jun 3 '17 at 3:49
  • $\begingroup$ No.. I dont know..can u please explain $\endgroup$ – NBN Jun 3 '17 at 3:50
  • $\begingroup$ @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: $p=-\alpha-\beta-\gamma$. $\endgroup$ – Lord Shark the Unknown Jun 3 '17 at 3:55
  • $\begingroup$ Hint: Look up Vieta's formula for the sum of the roots of a polynomial. $\endgroup$ – John Wayland Bales Jun 3 '17 at 4:00

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 $$a^3+b^3+c^3=(1/15)((-7a+11)+(-7b+11)+(-7c+11))=33/15=11/5.$$


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