Question

If `(2a)/(3b) = (3b)/(4c)`, show that (2a – 3b + 4c) (2a + 3b + 4c) = 4a² + 9b² + 16c².

Answer 1

Given: `(2a)/(3b) = (3b)/(4c)`

We want to show (2a – 3b + 4c)(2a + 3b + 4c) = 4a² + 9b² + 16c²

Stepwise calculation:

1. From the given equation:

\[ \frac{2a}{3b} = \frac{3b}{4c} \] 

Cross-multiplied:

(2a)(4c) = (3b)(3b) 

⇒ 8ac = 9b²

2. Expand the left-hand side (L.H.S.) using the formula (x – y)(x + y) = x² – y², where here the terms are grouped carefully: 

(2a – 3b + 4c)(2a + 3b + 4c)

We can consider A = 2a, B = 3b, C = 4c, then (A – B + C)(A + B + C) = (A + C)² – B² 

Expanding, (A² + 2AC + C²) – B² = A² + C² – B² + 2AC

Putting back (A, B, C):

= (2a)² + (4c)² – (3b)² + 2 × 2a × 4c

= 4a² + 16c² – 9b² + 16ac

3. Recall from Step 1 that 8ac = 9b²

⇒ 16ac = 2 × 9b²

⇒ 16ac = 18b² 

Substitute 16ac = 18b² into the expression above: 

4a² + 16c² – 9b² + 16ac = 4a² + 16c² – 9b² + 18b²

4a² + 16c² – 9b² + 16ac = 4a² + 16c² + 9b²

Rearranging the terms:

4a² + 9b² + 16c²