Question
If $a+b+c=9$ and $a b+b c+c a=40$, find $a^2+b^2+c^2$.
✓
Answer
Recall the formula
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$(a+b+c)=9, a b+b c+c a=40$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(9)^2=a^2+b^2+c^2+2 \cdot(40)$
$a^2+b^2+c^2+80=81$
$a^2+b^2+c^2=81-80$
$a^2+b^2+c^2=1$
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
Given that
$(a+b+c)=9, a b+b c+c a=40$
Then we have
$(a+b+c)^2=a^2+b^2+c^2+2(a b+b c+c a)$
$(9)^2=a^2+b^2+c^2+2 \cdot(40)$
$a^2+b^2+c^2+80=81$
$a^2+b^2+c^2=81-80$
$a^2+b^2+c^2=1$
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