the equation (x-a)^3+(x-b)^3+(x-c)^3=0 (a≠b≠c) has
a.all roots equal
b.one real and 2 imaginary
c.three real roots namely x=a,b,c
d.none

Question

the equation (x-a)^3+(x-b)^3+(x-c)^3=0 (a≠b≠c) has
a.all roots equal
b.one real and 2 imaginary
c.three real roots namely x=a,b,c
d.none

anonymous · Answer

c

anonymous · Answer

observation!!! :D

rsadhvika · Answer

x=a will not give u a zero

rsadhvika · Answer

since its a cubic, it must have one real root for sure

samigupta8 · Answer

sorry it's b option that is correct

rsadhvika · Answer

expand it, and apply descartes rule of signs

rsadhvika · Answer

or you can look at options, and eliminate the options that make no sense. this wud be easy here

samigupta8 · Answer

$$x^3-a^3-3x^2a+3xa^2+x^3-b^3-3x^2b+3xb^2+x^3-c^3-3x^c+3xc^2$$

rsadhvika · Answer

elimination would be easy here, as we dont know the signs of a, b, c... so applying descartes ends up messy wid many cases

rsadhvika · Answer

First see that, given equation is a polynomial of degree 3
so it must have atleast 3 distinct zeros. So strike off  last option

rsadhvika · Answer

see my first reply -- you can strike off option c also

samigupta8 · Answer

first option wud also be discarded since a is not equal to b is not equal to c and hence we cannot hva e3 real zeroes too 
...........ryt

rsadhvika · Answer

partially correct.

first option can be discarded because :-  since the given equation is a cubic, to have same 3 real zeroes, you need to bring the equation to form : $(x-k)^3 = 0$. 

since a, b, c are different, the given equaiton can never become a perfect cube

directrix · Answer

@divu.mkr Code of conduct: OpenStudy values the Learning process - not the ‘Give you an answer’ process Don’t post only answers - guide the asker to a solution http://openstudy.com/code-of-conduct