If the roots of $x^3 − 6x^2 + 10x + 1$ are denoted as a, b, c, then find the value of $(a^2 + b^2 )(a^2 + c^2 )(b^2 + c^2 )$.

Question

If the roots of $x^3 − 6x^2 + 10x + 1$ are denoted as a, b, c, then find the value of $(a^2 + b^2 )(a^2 + c^2 )(b^2 + c^2 )$.

I have tried factoring $x^3 − 6x^2 + 10x + 1$ but didn't get anything. Can someone let me know how to start solving? Thank you!

Answer 1

Hint $1:$ Using the Vieta's formula, you can find: $$(x+y)(y+z)(z+x) = (x+y+z)(xy+yz+zx)-xyz$$ if you knew what equation $x,y,z$ solves.

Hint $2:$ Then if you knew what $x,y,z$ solves, can you find what equation $x^2,y^2,z^2$ by using the Vieta formulae again?

Answer 2

Alt. hint: $\;$ the polynomial can be written as:

$$x^3 − 6x^2 + 10x + 1 = (x-a)(x-b)(x-c)$$

Substituting $x \mapsto -x\,$:

$$x^3 + 6x^2 + 10x - 1 = (x+a)(x+b)(x+c)$$

Multiplying the two:

$$(x^3+10x)^2-(6x^2-1)^2 = (x^2-a^2)(x^2-b^2)(x^2-c^2)$$

Expanding the LHS and substituting $x^2 \mapsto t\,$:

$$p(t) = t^3 - 16 t^2 + 112 t - 1 = (t-a^2)(t-b^2)(t-c^2)$$

The sum of the roots in $t$ is $s = a^2+b^2+c^2 = 16$ by Vieta's relations, and the expression being sought is $(a^2 + b^2 )(a^2 + c^2 )(b^2 + c^2 )=(s-a^2)(s-b^2)(s-c^2) = p(s)$.

Answer 3

Just another way, with $p(x) = x^3-6x^2+10x+1$,
let $s^2=a^2+b^2+c^2 = (a+b+c)^2-2(ab+bc+ca) = 6^2-2\cdot 10 =16 \implies s = \pm4$. Then what we seek is:$$(s^2-a^2)(s^2-b^2)(s^2-c^2) = \prod_{cyc} (s-a) \cdot \prod_{cyc} (s+a) = p(s)\cdot \left(-p(-s)\right) = -p(4)\cdot p(-4)$$

Answer 4

Just enjoying algebra... and answering questions

define

$x^3+px^2+qx+r=0\qquad$it is well known that $\quad p=-(a+b+c),\quad q=ab+ac+bc,\quad r=-abc$

$p_2=a^2+b^2+c^2=(a+b+c)^2-2(ab+ac+bc)=(-p)^2-2q$

then

$$\begin{aligned} (a^2+b^2)(a^2+c^2)(b^2+c^2)&=(p_2-a^2)(p_2-b^2)(p_2-c^2)\\ &=p_2^3-(a^2+b^2+c^2)p_2^2+(a^2b^2+a^2c^2+b^2c^2)p_2-(abc)^2\\ &=p_2^3-p_2^3+Xp_2-(abc)^2\\ &=X(p^2-2q)-r^2 \end{aligned}$$ this $\quad X:\quad$ $$\begin{aligned}X&=(ab)^2+(ac)^2+(bc)^2\\ &=(ab+ac+bc)^2-2(ab\cdot ac+ab\cdot bc+ac\cdot bc)\\ &=q^2-2(abc)(a+b+c)\\ &=q^2+2r(-p) \end{aligned}$$

Then the quantity sought is

$(q^2-2rp)(p^2-2q)-r^2\quad$ which evaluates to $\quad 1791\quad$ if $\quad (p,q,r)=(-6,10,1)$

Answer 5

We have $e_1=6,e_2=10,e_3=-1$. Then, $$\begin{array} .(a^2+b^2)(b^2+c^2)(c^2+a^2)&=(p_2-c^2)(p_2-a^2)(p_2-b^2)\\ &=p_{2}^3-p_{2}^3+(a^2b^2+b^2c^2+c^2a^2)p_2-a^2b^2c^2\\ &=(e_{2}^2-2e_1e_3)(e_1^2-2e_2)-e_3^2\\ &=(10^2-2(-1)6)(6^2-20)-(-1)^2\\ &=112\times 16-1 =1791 \end{array}$$