They are all related

The problem can be solved using Newton's Sums method.

$\begin{aligned} \Rightarrow a^2+b^2+c^2 & = (a+b+c)^2 - 2(ab+bc+ca) \\ \lambda & = \lambda^2 - 2(ab+bc+ca) \\ \Rightarrow ab+bc+ca & = \frac {\lambda^2 - \lambda}{2} \\ \Rightarrow a^3+b^3+c^3 & = (a+b+c)(a^2+b^2+c^2) - (ab+bc+ca)(a+b+c) + 3abc \\ \lambda & = \lambda^2 - \left( \frac {\lambda^2 - \lambda}{2} \right) \lambda + 3(5!) \\ 2\lambda & = 2\lambda^2 - \lambda^3 + \lambda^2 + 720 \end{aligned}$

$\begin{aligned} \Rightarrow \lambda^3 - 3\lambda^2 + 2\lambda - 720 & = 0 \\ (\lambda -10)(\lambda^2 + 7\lambda + 72) & = 0 \\ \Rightarrow \lambda & = \boxed{10} \end{aligned}$

Let $a, b, c$ be roots of the cubic ${x}^{3}-s{x}^{2}+qx-p$ . Then, we have that $s=a+b+c, q=ab+bc+ac, p=abc$ . These may be used to express the system in another way:

$\displaystyle \begin{cases} { a }^{ 1 }+{ b }^{ 1 }+{ c }^{ 1 }=s \\ { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }={s}^{2}-2q \\ { a }^{ 3 }+{ b }^{ 3 }+{ c }^{ 3 }={s}^{3}-3qs+3p \end{cases}$

We know $\lambda=s$ , so $\lambda = {\lambda}^{2}-2q$ , therefore $q=\frac { { \lambda }^{ 2 }-\lambda }{ 2 }$ . Plugging this in, we have that $\displaystyle \lambda ={ \lambda }^{ 3 }-3\lambda \frac { { \lambda }^{ 2 }-\lambda }{ 2 } +3p$ so $p=\frac { \lambda { -\lambda }^{ 3 }+3\lambda \frac { { \lambda }^{ 2 }-\lambda }{ 2 } }{ 3 } =p$ . Simplifying, $6p={ \lambda }^{ 3 }-3{ \lambda }^{ 2 }+2\lambda$ . Note that $p=abc$ , so $p=5!$ and we must now solve $\displaystyle { \lambda }^{ 3 }-3{ \lambda }^{ 2 }+2\lambda -720=0$ Factor as $(\lambda-n)({\lambda}^{2}+k\lambda+\frac { 6! }{ n } )$ , and note that $n-k=3$ . We may assume that $n$ and $k$ are fairly close now. Notice the coefficient of $x$ is given by $\frac { 6! }{ n } -nk=2$ . (Again, we may use that this is small)

If we factor smart, we can use that $n$ and $k$ are fairly close and say that $\frac { 6! }{ n } \approx { n }^{ 2 }$ , therefore $n\approx \sqrt [ 3 ]{ 6! } \approx 9$ . With a little guess and check for integers around 9 (remember, $\lambda \in \mathbb{{Z}^{+}}$ ) we find $n=10$ . Thus, the real value of $\lambda$ is $10$ , as the quadratic has no real roots.

This problem can be solved by using this useful equation: (a+b+c)(a^2+b^2+c^2-ab-bc-ac)=a^3+b^3+c^3-3abc by using the conditions, we have x(x-(ab+bc+ac))=x-360 [1]

To solve for ab+bc+ac, we use this equation: 〖(a+b+c)〗^2=(a^2+b^2+c^2+2(ab+bc+ac)) again, by using the conditions, we'll get x^2=x+2(ab+bc+ac) Eventually, we'll come up with ab+bc+ac= (x^2-x)/2 plugging this in to [1], we'll get x^3-3x^2+2x-720=0 transposing 720 to the other side of the equation, x^3-3x^2+2x= 720 x(x^2-3x+2)=720 x(x-1)(x-2)=720

The only positive integer x which satisfies the equation is 10. Hence, x=10

A bit of trial and error* gives us this identity: $\frac16(a+b+c)^3-\frac12(a+b+c)(a^2+b^2+c^2)+\frac13(a^3+b^3+c^3)=abc$ Thus, substituting in our values, we get $\frac16\lambda^3-\frac12\lambda^2+\frac13\lambda=120$ , which can be rearranged into $\lambda(\lambda-1)(\lambda-2)=720$ . It doesn't take long to find $10$ as the solution.

*By trial-and-error, I basically mean "expanding out $(a+b+c)^3$ , $(a+b+c)(a^2+b^2+c^2)$ , and $(a^3+b^3+c^3)$ — those being the only degree-three equations I could get out of the expressions for $\lambda$ — and trying to manipulate them into $abc$ .

I used the same method as Chew-Seong Cheong , but with Newton's identity.