MCQ
If $a + b + c = 9$ and $ab + bc + ca =23,$ then $a^3+ b^3+ c^3 - 3abc =$
- ✓$108$
- B$207$
- C$669$
- D$729$
✓
Answer
Correct option: A.
$108$
Given, $a+b+c=9$
Hence, $(a+b+c)^2=81$
So, $a^2+b^2+c^2+2 a b+2 b c+2 c a=81$
i.e. $a^2+b^2+c^2+2(a b+b c+c a)=81$
i.e. $a^2+b^2+c^2+2(23)=81$
i.e. $a^2+b^2+c^2=81-46=35$
Now, $a^3+b^3+c^3-3 a b c$
$=(a+b+c)\left(a^2+b^2+c^2-a b-b c-c a\right)$
$=(a+b+c)\left[\left(a^2+b^2+c^2\right)-(a b+b c+c a)\right]$
$=(9)[35-23]$
$=9 \times 12$
$=108$
Hence, correct option is $(a).$
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