Question

prove that a square minus b square whole cube plus b square minus c square whole cube + c square minus A square hole cube equal to 3 a+ b* B + c * c plus a *a minus b * B minus c * c - a
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Answer 1

Answer:

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

Step-by-step explanation:

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

Numerator

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

using the fact

if x+y+z = 0, then x³+y³+ z³ = 3 x y z

x = a² - b²

y = b² - c²

z = c² - a²

x + y + z = 0  ( as a² - b² +b² - c² + c² - a²) = 0

= 3 (a² - b²) (b² - c²) (c² - a²)

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

similarly denominator =

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

as here x = a-b  , y = b-c , z = c-a    & x + y + z = 0

numerator/Denominator

= 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)

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