Question

$(x - y) 3 +( y - z)3 +( z - x)3 \div (x - y)(y - z)(z - x)$


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Answer 1

We know the corollary: if a+b+c=0 then a3+b3+c3=3abc

Using the above identity taking a=x−y, b=y−z and c=z−x, we have a+b+c=x−y+y−z+z−x=0 then the equation (x−y)3+(y−z)3+(z−x)3 can be factorised as follows:

(x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)

Hence, (x−y)3+(y−z)3+(z−x)3=3(x−y)(y−z)(z−x)

HOPE UH UNDERSTAND

Answer 2

Answer:

3x-3y+3y-3z+3z-3xdivided by(x+y) (y-z) (z-x)

albhabet is cut and

answer is 0.