Question

If a : b be the triplicate ratio of (a + x) : (b + x), prove that x³ − 3abx − ab(a + b) = 0.

Answer 1

Step 1: Set up the equation

`a/b = ((a + x)/(b + x))^3`

`a/b = (a + x)^3/(b + x)^3`

a(b + x)³ = b(a + x)³

Step 2: Expand both cubes

Expand the cubic terms using the identity

(p + q)³ = p³ + 3p²q + 3pq² + q³

`a/b = (a^3 + 3a^2x + 3ax^2 + x^3)/(b^3 + 3b^2x + 3bx^2 + x^3)`

Cross-multiply the terms:

a(b³ + 3b²x + 3bx² + x³) = b(a³ + 3a²x + 3ax² + x³)

ab³ + 3ab²x + 3abx² + ax³ = ba³ + 3ba²x + 3bax² + bx³

Step 3: Rearrange and simplify the equation

Move all terms to one side of the equation and group by powers of x:

(bx³ − ax³) + (3bax² − 3abx²) + (3ba²x − 3ab²x) + (ba³ − ab³) = 0

Simplify the coefficients:

(b − a)x³ + (3abx² − 3abx²) + (3a²bx − 3ab²x) + (a³b − ab³) = 0

(b − a)x³ + 0 + 3ab(a − b)x + ab(a² − b²) = 0

(b − a)x³ + 3ab(a − b)x + ab(a − b)(a + b) = 0

Step 4: Divide by the common factor (b − a)

Assuming a ≠ b, divide the entire equation by the common factor (b − a).

Note that (a − b) = −(b − a).

`((b - a)x^3)/(b - a) + (3ab(-b-a)x)/(b - a) + (ab(-(b - a))(a + b))/(b - a) = 0/(b - a)`

x3 − 3abx − ab(a + b) = 0