Newton's Sums

Let $S_1 = a+b+c$ , $\space S_2 = ab+bc+ca$ , $\space S_3 = abc\space$ and $\space P_n = a^n+b^n+c^n$ .

Then the value,

$S = a^2(1+a^2) + b^2(1+b^2) + c^2(1+c^2)$

$= a^2 + b^2 + c^2 + a^4 + b^4 + c^4 = P_2 + P_4$

Then using Vieta's Formula - Higher Degrees , we have: $S_1 = -3$ , $\space S_2 = 4$ , $\space S_3 = 8 \space$ . And using Newton's Identities we have:

$P_1 = S_1 = -3$

$P_2 = S_1P_1 - 2S_2 = (-3)(-3) -2(4) = 9 - 8 = 1$

$P_3 = S_1P_2 - S_2P_1 + 3S_3 = (-3)(1) -(4)(-3) + 3(8)$

$\quad \quad = -3+12+24 = 33$

$P_4 = S_1P_3 - S_2P_2 + S_3P_1 = (-3)(33) -(4)(1) + (8)(-3)$

$\quad \quad = -99-4-24 = -127$

Therefore, $S = P_2 + P_4 = 1 - 127 = \boxed {-126}$

It could easily be identified that $p(1)=0$

So $(x-1)$ is a factor of $p(x)$

$p(x)=(x-1)(x^{2}+4x+8)$

So other two roots will be $-2 \pm 2i$

So $a=1 \ ,\ b=-2 + 2i \ ,\ c=-2 - 2i$ where $i= \sqrt{-1}$

The given expression can be written as

$2 + -8i(1-8i)+8i(1+8i)$

$=2-128$

So the answer is $\boxed{-126}$

$\color{#69047E}{a^2(1+a^2)+b^2(1+b^2)+c^2(1+c^2)=(a^2+b^2+c^2)+(a^4+b^4+c^4)}$ Define the following sums: $\color{#3D99F6}{\begin{cases}P_1=a+b+c\\P_2=a^2+b^2+c^2\\P_3=a^3+b^3+c^3\\P_4=a^4+b^4+c^4\end{cases}}$ Newton's sums tell us that: $\color{#20A900}{\begin{cases}P_1+3=0\\ P_2+3P_1+8=0 \\ P_3+3P_2+4P_1-24=0 \\ P_4+3P_3+4P_2-8P_1=0 \end{cases}}$ Solving this system of equations,we get $\color{#D61F06}{P_1=-3,P_2=1,P_3=33,P_4=-127}$ .Out of these, $\color{#D61F06}{P_2\;and\;P_4}$ are the ones we need.So $\color{#EC7300}{(a^2+b^2+c^2)+(a^4+b^4+c^4)=P_2+P_4=1+(-127)=1-127=\boxed{-126}}$

Please note that the expression in the question can also be written as $\color{#3D99F6}a^4+b^4+c^4+a^2+b^2+c^2$

Let's begin!

If $a,b,c$ are the roots, then

$a^3+3a^2+4a-8=0$ ... $(i)$

$b^3+3b^2+4b-8=0$ ... $(ii)$

$c^3+3c^2+4c-8=0$ ... $(iii)$

Multipying $a$ to $(i)$ , $b$ to $(ii)$ and $c$ to $(iii)$ , we get

$\color{#3D99F6}a^4+\color{#D61F06}3a^3+\color{#3D99F6}4a^2-\color{#D61F06}8a$ $=0$

$\color{#3D99F6}b^4+\color{#D61F06}3b^3+\color{#3D99F6}4b^2-\color{#D61F06}8a$ $=0$

$\color{#3D99F6}c^4+\color{#D61F06}3c^3+\color{#3D99F6}4c^2-\color{#D61F06}8a$ $=0$

Adding the above equations, we get

$\color{#3D99F6}a^4+b^4+c^4+\color{#D61F06}3(a^3+b^3+c^3)+\color{#3D99F6}4(a^2+b^2+c^2)\color{#D61F06}-8(a+b+c)$ $=0$ .

$\implies$ $\color{#3D99F6}{(a^4+b^4+c^4+a^2+b^2+c^2)}+\color{#D61F06}3(a^3+b^3+c^3)+\color{#3D99F6}3(a^2+b^2+c^2)\color{#D61F06}-8(a+b+c)$ $=0$ .

$\implies$ $\color{#3D99F6}(a^4+b^4+c^4+a^2+b^2+c^2)=\color{#D61F06}-3(a^3+b^3+c^3)\color{#3D99F6}-3(a^2+b^2+c^2)\color{#D61F06}+8(a+b+c)$

Hence, we have arrived at an equation for the required expression. All we need to do is to plug in the values of $\color{#D61F06}(a^3+b^3+c^3)$ , $\color{#3D99F6}(a^2+b^2+c^2)$ and $\color{#D61F06}(a+b+c)$ .

By Vieta's Formula we have

$\color{#D61F06}a+b+c$ $=-3$

$ab+bc+ac=4$

$abc=8$

Let's find the values of the unknowns...

Firstly,

$\color{#3D99F6}a^2+b^2+c^2$ $=(a+b+c)^2 -2(ab+bc+ac)$

$\color{#3D99F6}a^2+b^2+c^2=$ $(-3)^2 -2(4)=\color{#3D99F6}1$

Secondly we have,

$\color{#D61F06}a^3+b^3+c^3$ $=(a+b+c)[a^2+b^2+c^2-(ab+bc+ac)]+3abc$

$\color{#D61F06}a^3+b^3+c^3$ $=(-3)[1-4]+3(8)=\color{#D61F06}33$

Therefore,

$\color{#3D99F6}(a^4+b^4+c^4+a^2+b^2+c^2)$ $=\color{#D61F06}-3(33)-\color{#3D99F6}3(1)+\color{#D61F06}8(-3)=\color{#69047E}\boxed{-126}$ .

apply Vieta's formulas

a^2(1 + a^2) + b^2(1 + b^2) + c^2(1 + c^2)

= a^2 + b^2 + c^2 + a^4 + b^4 + c^4

a, b, and c are roots

We'll use these values later in the other equations

a + b + c = -3 (second numerical coefficient over the leading coefficient)

abc = 8 (last numerical coefficient over the leading coefficient)

ab + ac + bc = 4 (third numerical coefficient over the leading coefficient)

(a + b + c)^2 = (-3)^2 = 9

(a + b + c)^2 = 9

=> a^2 + b^2 + c^2 + 2(ab + ac + bc) = 9

=> a^2 + b^2 + c^2 + 2(4) = 9

=> a^2 + b^2 + c^2 = 1 <--- will be used on the main problem

(ab + ac + bc)^2 = (4)^2 = 16

(ab + ac + bc)^2 = 16

=> a^2b^2 + a^2c^2 + b^2c^2 + 2abc(a + b + c) = 16

=> a^2b^2 + a^2c^2 + b^2c^2 + 2(8)(-3) = 16

=> a^2b^2 + a^2c^2 + b^2c^2 - 48 = 16

=> a^2b^2 + a^2c^2 + b^2c^2 = 64 <-- will be used in the next equation

(a^2 + b^2 + c^2)^2 = (1)^2 = 1

(a^2 + b^2 + c^2)^2 = 1

=> a^4 + b^4 + c^4 + 2(a^2b^2 + a^2c^2 + b^2c^2)=1

=> a^4 + b^4 + c^4 + 2(64) = 1

=> a^4 + b^4 + c^4 = -127 <-- will be used on the main problem

going back to the main problem:

find the value of

a^2 + b^2 + c^2 + a^4 + b^4 + c^4

simply just substitute:

a^2 + b^2 + c^2 = 1

a^4 + b^4 + c^4 = -127

therefore

a^2 + b^2 + c^2 + a^4 + b^4 + c^4

= 1 -127

= -126

Notice x=1 is a solution so factor x^3+3x^2+4x-8 to (x-1)(x^2+4x+8). let a=1 and the roots of x^2+4x+8 be b and c. (x-b)(x-c)= x^2+4x+8. so bc=8 and b+c=-4. expand and simplify a^2(1+a^2) + b^2(1+b^2) +c^2(1+c^2) with a=1. so then we have 2+b^2+c^2+c^4+b^4. solve for a^2+b^2 with b+c=-4 and bc=8 (b+c)^2=b^2+c^2+2*bc. plug in value of bc and b+c. so (-4)^2=b^2+c^2+2(8). Then b^2+c^2=0. Then solve for b^4+c^4. (b^2+c^2)^2 =b^4+c^4+ 2(bc)^2. 0^2 =b^4+c^4+2(8)^2 so b^4+c^4=-128. So the value we want is 2+0-128 =-126