Question

If $a+b+c=9$ and $ab+bc+ca=26$, then the value of $a^2+b^2+c^2-3$ is _______.

• A. 25
• B. 29
• C. 28
• D. 26

Answer 1

The correct option is D 26

Given:
$a+b+c=9ab+bc+ca=26$

Squaring on both sides of $a+b+c=9$, we get
$(a+b+c{)}^2=81$

We know that, $(a+b+c{)}^2=a^2+b^2+c^2+2(ab+bc+ca)$]
$81=a^2+b^2+c^2+2(ab+bc+ca)$
Substituting the value of $ab+bc+ca=26$ in the above expression,
$81=a^2+b^2+c^2+2\left(26\right)$
$=81–52=a^2+b^2+c^2=29$

$a^2+b^2+c^2-3=29-3=26$