How to Evaluate $\dfrac{a^5+b^5+c^5-1}{abc}$ Given $ab+bc+ca=0$ and $abc\ne 0$?

[anemone]

Let $a,\,b,,c \in \mathbb R$ such that $ab+bc+ca=0$ and $abc\ne 0$. Evaluate $\dfrac{a^5+b^5+c^5-1}{abc}$.

 

[jvicens]

Hello,

Interesting problem! First, let's rewrite the given conditions as $ab = -bc - ca$ and $abc \neq 0$. We can then substitute this into the expression we are trying to evaluate to get:

$\dfrac{a^5+b^5+c^5-1}{abc} = \dfrac{a^5 + (-bc)^5 + (-ca)^5 - 1}{abc}$

Using the binomial theorem, we can expand the terms with exponents to get:

$\dfrac{a^5 + b^5 + c^5 - 1}{abc} = \dfrac{a^5 - b^5 - c^5 + 1}{abc}$

Next, we can factor the numerator as a difference of squares:

$\dfrac{a^5 - b^5 - c^5 + 1}{abc} = \dfrac{(a^2-b^2)(a^3+b^3) - c^5 + 1}{abc}$

Using the identities $a^2-b^2 = (a-b)(a+b)$ and $a^3+b^3 = (a+b)(a^2-ab+b^2)$, we can further simplify the numerator to get:

$\dfrac{(a-b)(a+b)(a^2-ab+b^2)(a^2+b^2) - c^5 + 1}{abc}$

Since $ab = -bc - ca$, we can rewrite $a^2 + b^2$ as $(a+b)^2 - 2ab$. Substituting this in, we get:

$\dfrac{(a-b)(a+b)(a^2-ab+b^2)((a+b)^2 - 2ab) - c^5 + 1}{abc}$

Using the fact that $ab = -bc - ca$ again, we can simplify the expression further to get:

$\dfrac{(a-b)(a+b)(a^2-ab+b^2)(a^2 + b^2 + 2bc + 2ca) - c^5 + 1}{abc}$

Finally, we can substitute $ab = -bc - ca$ and $abc \neq 0$ into the expression to get our final answer of:

$\dfrac{(a-b)(a+b)(a^2-ab+b^2)(a^