Question

Without cubing simplify the following
(a²-b²)³ + (b²-c²)³ + (c²-a²)³.​

Answer 1

Step-by-step explanation:

(a2−b2)3+(b2−c2)3+(c2−a2)3(a−b)3+(b−c)3+(c−a)3(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=(a2−b2),y=(b2−c2),z=(c2−a2)

⟹x+y+z=0(1)(1)⟹x+y+z=0

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

0=x3+y3+z3+3(0)(xy+yz+zx)−3xyz0=x3+y3+z3+3(0)(xy+yz+zx)−3xyz

Answer 2

Answer:

$\frac{(a2−b2)3+(b2−c2)3+(c2−a2)3}{(a−b)3+(b−c)3+(c−a)3</p><p> } \\ \\ Let \: in \: numerator \\ \\ \:x = ( {a}^{2} - {b}^{2} ) \: y = { b}^{2} - ( {c}^{2} ) \: z = ({c}^{2} - {a}^{2} )$

$= > x + y + z = 0 \: \: \: \: \: \: \: |1 \\ (x + y + z)^{3} = {x }^{3} + {y}^{3} + {z}^{3} \: \\ we \: know. + 3(x + y + z)(xy + yz + zx) - 3xyz \\ 0 = {x}^{3} + {y}^{3} + {z}^{3} + 3(0)(xy + yz + zx) - 3xyz$