Question

Simplify : (a²-b²)³ + (b²-c²)³ + (c²-a²)³/(a-b)³ + (b-c)³ + (c-a)³​

Answer 1

Answer:

we use the fact that if a+b+c = 0, then a³+b³+ c³ = 3 a b c.

(a² - b²) + (b² - c²) + (c² - a²) = 0...

also (a-b) + (b-c)+ (c-a) = 0

we assume that a ≠b ≠ c.

hence,

(a²-b²)³+(b²-c²)³+(c²-a²)³ ÷ (a-b)³+b-c)³+(c-a)³

= 3 (a² - b²) (b² - c²) (c² - a²) / [ 3 (a - b) (b - c) ( c -a) ]

= ( a+b) (b + c) ( c +a)

∵ a² - b² = (a -b) (a+b) and similarly other term

Answer 2

Answer:

a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3

Let in numerator

x=(a2−b2),y=(b2−c2),z=(c2−a2)

⟹x+y+z=0(1)

We know, (x+y+z)3=x3+y3+z3+3(x+y+z)(xy+yz+zx)−3xyz

0=x3+y3+z3+3(0)(xy+yz+zx)−3xyz

x3+y3+z3=3xyz(2)

Let in denominator,

p=(a−b),q=(b−c),r=(c−a)

⟹p+q+r=0

⟹

p3+q3+r3=3pqr(3)

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3=3xyz3pqr

=(a2−b2)(b2−c2)(c2−a2)(a−b)(b−c)(c−a)

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

=(a+b)(b+c)(c+a)