If $$a^3+b^3:a^3-b^3=185:158$$​ then $$a:b=?$$​​

Correct option is B

Given:

​$$\frac{a^3 + b^3}{a^3 - b^3} = \frac{185}{158}$$​

Find a : b 

Concept Used:
Apply componendo and dividendo:
If$$\frac{A}{B} = \frac{m}{n}$$​, then

​$$\frac{A + B}{A - B} = \frac{m + n}{m - n}$$​​

Solution:
$$\frac{a^3 + b^3 + a^3 - b^3}{a^3 + b^3 - (a^3 - b^3)} = \frac{185 + 158}{185 - 158}$$​​

​$$\frac{2a^3}{2b^3} = \frac{a^3}{b^3} = \frac{343}{27}$$​​

So,$$\frac{a}{b} = \frac{7}{3}$$

​Alternate Method:​

Let a : b = x : 1

a = x, b = 1
Substitute in the given ratio:

​$$\frac{x^3 + 1}{x^3 - 1} = \frac{185}{158}$$​

​$$158(x^3 + 1) = 185(x^3 - 1)$$​

​$$158 + 185 = 185x^3 - 158x^3$$​

​$$343 = 27x^3$$​

​$$x^3 = \frac{343}{27} = \left(\frac{7}{3}\right)^3$$​​

​$$x = \frac{7}{3}$$​

So,$$a : b = x : 1 = \frac{7}{3} : 1 = 7 : 3$$