Giải thích các bước giải:
Ta có:
$\dfrac{a}{b}=\dfrac{c}{d}$
$\to\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2b}{2d}=\dfrac{3a}{3c}=\dfrac{5b}{5d}=\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}=\dfrac{3a+5b}{3c+5d}$
$\to\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}=\dfrac{3a+5b}{3c+5d}$
$\to(\dfrac{a}{c})^3=(\dfrac{b}{d})^3=(\dfrac{a+b}{c+d})^3=(\dfrac{a-2b}{c-2d})^3=(\dfrac{3a+5b}{3c+5d})^3$
$\to\dfrac{a^3}{c^3}=\dfrac{b^3}{d^3}=\dfrac{(a+b)^3}{(c+d)^3}=\dfrac{(a-2b)^3}{(c-2d)^3}=\dfrac{(3a+5b)^3}{(3c+5d)^3}$
