The correct option is A True
Formula used: if a+b+c=0, then a3+b3+c3=3abc
125(x−y)3+(5y−3z)3+(3z−5x)3=(5x−5y)3+(5y−3z)3+(3z−5x)3
Now, (5x−5y)+(5y−3z)+(3z−5x)=0
⇒(5x−5y)3+(5y−3z)3+(3z−5x)3=3(5x−5y)(5y−3z)(3z−5x)
=15(x−y)(5y−3z)(3z−5x)
∴125(x−y)3+(5y−3z)3+(3z−5x)3=15(x−y)(5y−3z)(3z−5x)