Question

SOLVE this problem attched


prove..​

Answer 1

Let , ( a^2 - b^2 ) = x  ---------------------- (i)

       ( b^2 - c^2 ) = y  ---------------------- (ii)

       ( c^2 - a^2 ) = z ----------------------- (iii)

Adding all these three equations,

a^2 - b^2 + b^2 - c^2  + c^2 - a^2 = x + y + z

0 = x + y + z

We know that when x + y + z = 0, then x³ + y³ + z³ = 3xyz

Now,

x³ + y³ + z³ = 3xyz

By the the values of x, y and z in the above equation,

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

     = 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) 

Hope this helps!

Answer 2

Answer:

Let , ( a^2 - b^2 ) = x  ---------------------- (i)        ( b^2 - c^2 ) = y  ---------------------- (ii)        ( c^2 - a^2 ) = z ----------------------- (iii)

Adding all these three equations,

a^2 - b^2 + b^2 - c^2  + c^2 - a^2

= x + y + z0

= x + y + z

We know that when x + y + z = 0,

then x³ + y³ + z³ = 3xyz

Now,x³ + y³ + z³ = 3xyz

By the the values of x, y and z in the above equation,

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

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