Let us start from the expression x_{1}^{3}+x_{1}^{2}+x_{1} and use that x_{1} is a root of the quadratic equation. Then we obtain that x_{1}^{2}(x_{1}+1)+x_{1}=(8x_{1}-11)(x_{1}+1)+x_{1}=
8x_{1}^{2}+8x_{1}-11x_{1}-11+x_{1}=8(8x_{1}-11)-2x_{1}-11=64x_{1}-88-2x_{1}-11=62x_{1}-99. Similarly we get that x_{2}^{3}+x_{2}^{2}+x_{2}=62x_{2}-99. Then by adding everything up we arrive at
x_{1}^{3}+x_{1}^{2}+x_{1}+x_{2}^{3}+x_{2}^{2}+x_{2}=62x_{1}-99+62x_{2}-99=62(x_{1}+x_{2})-198=
62(8)-198=496-198=298 where we used that from the Vieta's rules we have that x_{1}+x_{2}=8.
Nikolaos P.
tutor
Because x_{1} is a solution of the quadratic equation it must satisfy the equation, that is x_{1}^{2}=8x_{1}-11. So every time the quantity x_{1}^{2} appears we substitute it by 8x_{1}-11.
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11/06/20
Jack F.
Ok, I understand that part! Now, how did you get the 8x_{1} - 11? Sorry!
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11/06/20
Jack F.
Thank you for this information. But, how did you get to the second part? "then we obtain that. . ."11/06/20