Giải thích các bước giải:

Ta có:

$\dfrac{a}{b}=\dfrac{c}{d}$

$\to\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{2b}{2d}=\dfrac{3a}{3c}=\dfrac{5b}{5d}=\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}=\dfrac{3a+5b}{3c+5d}$

$\to\dfrac{a}{c}=\dfrac{b}{d}=\dfrac{a+b}{c+d}=\dfrac{a-2b}{c-2d}=\dfrac{3a+5b}{3c+5d}$

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

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

Ta có : `a/b=c/d`

`⇒a/c=b/d` `(1)`

`⇒(a/c)^3=(b/d)^3`

`=a^3/c^3=b^3/d^3` `(2)`

Từ `(1),⇒a/c=b/d=2b/2d=3a/3c=5b/5d`

Áp dụng tính chất dãy tỷ số bằng nhau

`⇒a/c=b/d={a-2b}/{c-2d}={3a+5b}/{3c+5d}`

`⇒(a/c)^3=(b/d)^3=({a-2b}/{c-2d})^3=({3a+5b}/{3c+5d})^3` `(3)`

Từ `(2)` và `(3),⇒a^3/c^3=b^3/d^3=({a-2b}/{c-2d})^3=({3a+5b}/{3c+5d})^3` `(đpcm)`