- #1
anemone
Gold Member
MHB
POTW Director
- 3,883
- 115
Let $a,\,b,\,c$ be integers such that $(a-b)^2+(b-c)^2+(c-a)^2=abc$. Prove that $a^3+b^3+c^3$ is divisible by $a+b+c+6$.
The formula for proving divisibility of $a^3+b^3+c^3$ using $(a-b)^2+(b-c)^2+(c-a)^2=abc$ is as follows: If $a$, $b$, and $c$ are integers, then $(a-b)^2+(b-c)^2+(c-a)^2$ must be divisible by $abc$.
The formula $(a-b)^2+(b-c)^2+(c-a)^2$ is derived from the expansion of $(a+b+c)^3$. It helps to simplify and manipulate the expression $a^3+b^3+c^3$ into a form that is easily divisible by $abc$.
The expression $(a-b)^2+(b-c)^2+(c-a)^2$ represents the sum of squares of the differences between any two of the three integers $a$, $b$, and $c$. This expression plays a crucial role in proving the divisibility of $a^3+b^3+c^3$ by $abc$.
Yes, the formula $(a-b)^2+(b-c)^2+(c-a)^2=abc$ can be used to prove the divisibility of other expressions, as long as those expressions can be manipulated into a form that is divisible by $abc$.
Yes, the formula $(a-b)^2+(b-c)^2+(c-a)^2$ can be interpreted geometrically as the sum of the squares of the lengths of the three sides of a triangle with side lengths $a-b$, $b-c$, and $c-a$. This can help visualize and understand the proof of divisibility of $a^3+b^3+c^3$ using $(a-b)^2+(b-c)^2+(c-a)^2=abc$.