Q13/055) Let a,b,c be roots of equation x^3−6x^2+kx+k=0 and (a−1)^3+(b−2)^3+(c−3)^3=0. Find a, b, c and k.

We have sum of the roots

a+b+c = 6

so (a + b+ c- 6) = 0

or (a-1)+(b-2)+(c-3) = 0

so (a-1)^3 + (b-2)^3 + (c-3)^3= 3(a-1)(b-2)(c-3) ( as x+ y + z = 0 => x^3+y^3 + z^3 = 3xyz

so 3(a-1)(b-2)(c-3) = 0 => a= 1 or b= 2 or c = 3

f(x) = x^3−6x^2+kx+k

  taking  1 as a root we get

f(x)=  1-6+ k+ k = 0 or k = 5/2

so we get x^3-6x^2 + 5/2 x + 5/2 = 0

or 2x^3 – 12x^2 + 5 x + 5 = 0

factoring we get (x-1)(2x^2 – 10 x - 5) = 0

we can solve (2x^2 – 10 x - 5)= 0 to get (10+/-√(140))/4 so b = (5+ √35)/2 and c = (5+ √35)/2 or b = (5-  √35)/2 and c = (5+ √35)/2  

similarly other 4 set of roots can be found