Question

If a, b, and c are the tribes of the geometry sequence, prove that:
(a + ab + b²) / (bc + ca + ab) = a / b

Answer 1

$\displaystyle \frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a^2+a\cdot ar+(ar)^2}{ar\cdot ar^2+ar^2\cdot a+a\cdot ar}\\\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a^2+ a^2r+a^2r^2}{a^2r^3+a^2r^2+a^2r}\\\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{1+r+r^2}{r^3+r^2+r}\\\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{1+r+r^2}{r(r^2+r+1)}\\\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{1}{r}\cdot\frac{a}{a}\\\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a}{ar}\\\boxed{\boxed{\frac{a^2+ab+b^2}{bc+ca+ab}=\frac{a}{b}}}$