Question

If $a+b+c=1;a^2+b^2+c^2=2;a^3+b^3+c^3=3$, then

• A. $ab+bc+ca=-\frac{1}{2}$
• B. $abc=\frac{1}{6}$
• C. $ab+bc+ca=\frac{1}{2}$
• D. $abc=\frac{1}{9}$

Answer 1

Answer: (A), (B)

The correct options are
A $ab+bc+ca=-\frac{1}{2}$
B $abc=\frac{1}{6}$

Given, $a+b+c=1$

$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)=1$

Squaring on both the side

$\Rightarrow ab+bc+ca=-\frac{1}{2}$

$\therefore a^3+b^3+c^3-3abc$

$=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$

$\Rightarrow 3-3abc=\left(1\right)(2-(-\frac{1}{2}))$

$\Rightarrow 3-2-\frac{1}{2}=3abc$

$\Rightarrow abc=\frac{1}{6}$