Please I need help on this question :

d) Show that $\beta_j^2=4-\alpha_j^2$ for $j\in\left\{1,2,3\right\}$ and conclude that $\beta_1^2\beta_2^2\beta_3^2=-f(2)f(-2)=7$ and hence $\beta_1\beta_2\beta_3=\sqrt{7}$ where $f(x)=x^3+x^2-2x-1$ and $\alpha_j=2\cos(2j\pi/7),$ $\beta_j=2\sin(2j\pi/7)$ for j = 1,2 and 3 I think for the first part you need to use the relation : $\cos^2(x) + \sin^2(x)= 1$ so $\beta_j^2=4-\alpha_j^2$ for $j\in\left\{1,2,3\right\}$ follows.

Thanks in advance


Note that $-\beta_j^2=(2-\alpha_j)(-2-\alpha_j)$ and so $-\beta_1^2\beta_2^2\beta_3^2=F(2)F(-2)$ where $F(x)=(x-\alpha_1)(x-\alpha_2)(x-\alpha_3)$. I presume you have already shown that this $F(x)$ is the same as your $f(x)$.

  • $\begingroup$ Thank you. yes I showed that f has three solutions $\alpha_1,\alpha_2,\alpha_3$ $\endgroup$ – Joe Nov 23 '17 at 3:13

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