Answer: 3(d−e)(e−f)(f−d)
Step by Step Explanation: - We know that
(a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca)…(i)
- Let us assume
(d−e)=a(e−f)=band (f−d)=c
We have+
(d−e)3+(e−f)3+(f−d)3=a3+b3+c3
Also,
(a+b+c)= (d−e)+(e−f)+(f−d)= d−e+e−f+f−d= 0
- Substituting (a+b+c)=0 in eq(i), we have
(a3+b3+c3−3abc)=(a+b+c)(a2+b2+c2−ab−bc−ca)⟹(a3+b3+c3−3abc)=0×(a2+b2+c2−ab−bc−ca)⟹(a3+b3+c3−3abc)=0⟹a3+b3+c3=3abc⟹a3+b3+c3=3(d−e)(e−f)(f−d)
- Hence, (d−e)3+(e−f)3+(f−d)3=3(d−e)(e−f)(f−d)