Question

Prove that : (a²- b²)³ + (b²- c²)³ + (c²- a²)³ = 3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)

It's a SA1 model question … Plz solve it…

Answer 1

Hey there!!

Given (a²-b²)³+(b²-c²)³+(c²-a²)³=3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)

Consider LHS:(a²-b²)³+(b²-c²)³+(c²-a²)³

If x + y + z =0 then x³ + y³+ z³ = 3xyz as,

Here x = a²-b², y = b²-c² and z = c²-a²

a²-b² + b²-c² + c²-a² = 0

Hence (a²-b²)³+(b²-c²)³+(c²-a²)³ = 3(a²-b²)(b²-c²)(c²-a²)³

= 3(a + b)(a − b)(b + c)(b − c)(c + a)(c − a)

= 3(a + b)(b + c)(c + a)(a − b)(b − c)(c − a)

∴ Hence proved

______________

Answer 2

Answer:

Step-by-step explanation:

Hey there!!

Given (a²-b²)³+(b²-c²)³+(c²-a²)³=3(a+b)(b+c)(c+a)(a-b)(b-c)(c-a)

Consider LHS:(a²-b²)³+(b²-c²)³+(c²-a²)³

If x + y + z =0 then x³ + y³+ z³ = 3xyz as,

Here x = a²-b², y = b²-c² and z = c²-a²

a²-b² + b²-c² + c²-a² = 0

Hence (a²-b²)³+(b²-c²)³+(c²-a²)³ = 3(a²-b²)(b²-c²)(c²-a²)³

= 3(a + b)(a − b)(b + c)(b − c)(c + a)(c − a)

= 3(a + b)(b + c)(c + a)(a − b)(b − c)(c − a)

∴ Hence proved