Question

Prove that (x+y)3 + (y+z)3 +(z+x)3 – 3(x+y)(y+z)(z+x) = 2(x3+y3+z3 -3xyz)

Answer 1

Answer:

here is it

Step-by-step explanation:

a3 + b3 + c3- 3abc = (a + b + c)(a2 + b2 + c2 – ab – bc – ca)

if a + b + c = 0; then a3 + b3 + c3- 3abc = 0

or a3 + b3 + c3= 3abc

For the numerator:

(x3 - y3) + (y3 - z3) + (z3 - x3) = 0

so, (x3 - y3)3 + (y3 - z3)3 + (z3 - x3)3

= 3 (x3 - y3) (y3 - z3) (z3 - x3)

again, x3 - y3= (x - y) (x2 + xy + y2)

so, 3 (x3 - y3) (y3 - z3) (z3 - x3)

= 3 (x - y) (x2 + xy + y2) (y - z) (y2 + yz + z2)(z - x) (z2 + zx + x2)

for the denominator:

(x - y) + (y-z) + (z-x) = 0

so, (x - y)3 + (y - z)3) (z - x)3 = 3 (x - y) (y-z) (z-x)

so, (x3 - y3)3 + (y3 - z3)3 + (z3 - x3)3 / (x - y)3 + (y - z)3) (z - x)3

= 3 (x - y) (x2 + xy + y2) (y - z) (y2 + yz + z2)(z - x) (z2 + zx + x2) / 3 (x - y) (y-z) (z-x)

= (x2 + xy + y2) (y2 + yz + z2) (z2 + zx + x2)