Question

The equation $(x-a{)}^3+(x-b{)}^3+(x-c{)}^3=0$ has

• A. all the roots are real
• B. one real and two imaginary
• C. 3 real roots namely $x=a,x=b,x=c$
• D. no real roots

Answer 1

Answer: (B)

one real and two imaginary

$(x-\alpha {)}^3=x^3-{\alpha }^3-3x^2\alpha +3x{\alpha }^2$
$f\left(x\right)=(x-a{)}^3+(x-b{)}^3+(x-c{)}^3=3x^3-3x^2(a+b+c)+3x(a^2+b^2+c^2)-a^3-b^3-c^3=0$
sum of roots $=a+b+c$
product of roots $=\frac{a^3+b^3+c^3}{3}$
product of roots taken two at a time$=a^2+b^2+c^2$
$f^{\prime }\left(x\right)=9x^2-6x(a+b+c)+3(a^2+b^2+c^2)$
$\Delta =b^2-4ac$
$=(a+b+c{)}^2-3(a^2+b^2+c^2)$
$=-(ab+bc+ca)<0$
$\therefore$ f' is not zero anywhere.
so f is increasing function, so it has one real and two imaginary roots.