Question

show that the equation (x-1)^3 +(x-2)^3 +(x-3)^3 +( x-4)^3 =0 has only one real root​

Answer 1

Answer:

(x−α)3=x3−α3−3x2α+3xα2

f(x)=(x−a)3+(x−b)3+(x−c)3=3x3−3x2(a+b+c)+3x(a2+b2+c2)−a3−b3−c3=0

sum of roots =a+b+c

product of roots =3a3+b3+c3

product of roots taken two at a time=a2+b2+c2

f′(x)=9x2−6x(a+b+c)+3(a2+b2+c2)

    Δ=b2−4ac

                =(a+b+c)2−3(a2+b2+c2)

               =−(ab+bc+ca)<0

∴ f' is not zero anywhere.

  so f is increasing function, so it has one real root