Evaluating the value of the roots of the expression

Question

I'm attempting this question below, I understand the method when its squared but not cubed. I evaluated $a^2 + b^2 + c^2$ as $(a + b + c)^2 -2(ab + ac + bc)$. I tried to obtain the answer by guess and check, changing a few of the powers around to try and work backwards but to no success. The answer for the question below is 23, and based upon the responses provided $a^4 + b^4 + c^4$ should be a breeze also? Thanks

Q: $x^3 - 2x^2 - 3x + 1 = 0$

$a^3 + b^3 +c^3 =?$

Answer 1

I suggest that you check this out: https://en.m.wikipedia.org/wiki/Newton%27s_identities

Now, the idea, as you guess, is to write $a^3+b^3+c^3$ as a function of the symmetric elementary polynomials.

Now, $a^3=2a^2+3a-1$, same for $b,c$.

Thus $a^3+b^3+c^3=2(a^2+b^2+c^2)+3(a+b+c)-3$, hence $a^3+b^3+c^3=2(a+b+c)^2+3(a+b+c)-4(ab+bc+ca)-3$.

Finally, $a^3+b^3+c^3=2*(2)^2+3*(2)-4*(-3)-3=23$.

Answer 2

$$x^3+1=2x^2+3x$$

Cubing both sides $$(x^3+1)^3=(2x^2+3x)^3=(2x^2)^3+(3x)^3+3(2x^2)(3x)(2x^2+3x)$$

Setting $x^3=y$ $$(y+1)^3=8y^2+27y+18y(y+1)$$

$$\iff y^3+y^2(3-8-18)+y(\cdots)+\cdots=0$$ whose roots are $a^3,b^3,c^3$

$$a^3+b^3+c^3=\dfrac{23}1$$

Answer 3

By Vieta's formulas: $$a+b+c=2, ab+bc+ca=-3, abc=-1.$$ Then you can use the formulas: $$\begin{align}A=a^2+b^2+c^2&=(a+b+c)^2-2(ab+bc+ca)=\\ &=2^2-2(-3)=\\ &=10.\end{align}$$ $$\begin{align}B=a^3+b^3+c^3&=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)+3abc=\\ &=(a+b+c)[(a+b+c)^2-3(ab+bc+ca)]+3abc=\\ &=2(2^2-3(-3))+3(-1)=\\ &=23.\end{align}$$ $$\begin{align}C=a^4+b^4+c^4&=(a^2+b^2+c^2)^2-2(a^2b^2+b^2c^2+c^2a^2)=\\ &=A^2-2[(ab+bc+ca)^2-2abc(a+b+c)]=\\ &=10^2-2[(-3)^2-2(-1)2]=\\ &=126.\end{align}$$ $$\begin{align}D=a^5+b^5+c^5&=(a+b+c)^5-\\ & \ \ \ \ \ \ 5(a+b)(b+c)(c+a)(a^2+b^2+c^2+ab+bc+ca)=\\ &=2^5-\frac{5[(a+b+c)^3-a^3-b^3-c^3]}{3}(a^2+b^2+c^2+ab+bc+ca)=\\ &=2^5-\frac{5[2^3-B]}{3}(A+(-3))=\\ &=2^5-\frac{5[-15]}{3}(7)=\\ &=207.\end{align}$$