An algebra problem

If:

$x = b + c$

$y = a + c$

$z = a + b$

then:

$\begin{aligned} &\frac{x^3+y^3+z^3-3xyz}{a^3+b^3+c^3-3abc} \\ &= \frac{(a+b)^3+(a+c)^3+(b+c)^3-3(a+b)(a+c)(b+c)}{a^3+b^3+c^3-3abc} \\ &= \frac{2(a^3+b^3+c^3)+3(a^{2}b+ab^2+a^{2}c+ac^{2}+b^{2}c+bc^{2})-3(a^{2}b+ab^2+a^{2}c+ac^{2}+b^{2}c+bc^{2}+2abc)}{a^3+b^3+c^3-3abc} \\ &= \frac{2(a^3+b^3+c^3-3abc)}{a^3+b^3+c^3-3abc} \\ &= 2 \end{aligned}$

as required.

Suponha que $\boxed{\mathsf{a = 2}}$ , $\boxed{\mathsf{b = 3}}$ e $\boxed{\mathsf{c = 4}}$ . Com efeito, tem-se que:

$\begin{cases} \mathsf{x = b + c \Rightarrow x = 3 + 4 \Rightarrow \boxed{\mathsf{x = 7}}} \\ \mathsf{y = a + c \Rightarrow y = 2 + 4 \Rightarrow \boxed{\mathsf{y = 6}}} \\ \mathsf{z = a + b \Rightarrow z = 2 + 3 \Rightarrow \boxed{\mathsf{z = 5}}}\end{cases}$

Desse modo, basta substituir na expressão...

$\mathsf{\frac{x^3 + y^3 + z^3 - 3xyz}{a^3 + b^3 + c^3 - 3abc} =}$

$\mathsf{\frac{7^3 + 6^3 + 5^3 - 3 \cdot 7 \cdot 6 \cdot 5}{2^3 + 3^3 + 4^3 - 3 \cdot 2 \cdot 3 \cdot 4} =}$

$\mathsf{\frac{54}{27} =} \\\\\\ \boxed{\boxed{\mathsf{2}}}$