Question

prove−(2ab+3c)2−(2ab−3c)2−24abc=0
​

Answer 1

Step-by-step explanation:

Solution :-

On taking LHS

(2ab+3c)² - (2ab-3c)² -24abc

We know that

(x+y)² = x2+2xy+y²

(2ab+3c)² = (2ab)²+2(2ab)(3c)+(3c)²

=> (2ab+3c)² = 4a²b²+12abc+9c²

and

(x-y)² = x²+2xy+y²

(2ab-3c)² = (2ab)²-2(2ab)(3c)+(3c)²

=> (2ab-3c)² = 4a²b²-12abc+9c²

Now,

(2ab+3c)² - (2ab-3c)² -24abc

=> (4a²b²+12abc+9c²)-(4a²b²-12abc+9c²)-24abc

=> 4a²b²+12abc+9c²-4a²b²+12abc-9c²-24abc

=> (4a²b²-4a²b²)+(9c²-9c²)+(12abc+12abc-24abc)

=> 0+0+24abc-24abc

=> 0

=> RHS

=> LHS = RHS

Alternative Method:-

On taking LHS

(2ab+3c)² - (2ab-3c)² -24abc

We know that

(x+y)²-(x-y)² = 4xy

Now,

(2ab+3c)² - (2ab-3c)² -24abc

=> 4(2ab)(3c) -24abc

=> 24abc-24abc

=> 0

=>RHS

=> LHS = RHS

Answer :-

(2ab+3c)² - (2ab-3c)² -24abc = 0

Used formulae:-

→ (x+y)² = x2+2xy+y²

→ (x-y)² = x²+2xy+y²

→ (x+y)²-(x-y)² = 4xy