Question

pls solve & get 100 pts​

Answer 1

If x+y+z = 0 then x³+y³+z³ = 3xyz */

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

After cancellation, we get

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

Answer 2

✨HOLA!!!✨

Let us take the numerator first,

$a - b + b - c + c - a$

$= 0$

Now , we know that if a + b + c = 0 , then a cube + b cube + c cube = 3abc .

So, Numerator = 3(a-b)(b-c)(c-a)

Now, let us take the denominator ,

${a}^{2} - {b}^{2} + {b}^{2} - {c}^{2} + {c}^{2} - {a}^{2}$

$= 0$

Now,we know that if a + b + c = 0 , then a cube + b cube + c cube = 3abc .

So, numerator becomes =

$3( {a}^{2} - {b}^{2} )( {b}^{2} - {c}^{2} )( {c}^{2} - {a}^{2} )$

So ,

$(3( {a}^{2} - {b}^{2} )( {b}^{2} - {c}^{2} )( {c}^{2} - {a}^{2} )) \div( 3(a - b)(b - c)(c - a)$

$= (((a + b)(a - b)) ((a + b)(a - b)) ((a + b)(a - b))) \div ((a - b)(b - c)(c - a))$

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

HOPE IT HELPS UHH #CHEERS