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$, a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 81$
$i.e. a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81$
$i.e. a^2 + b^2 + c^2 + 2(23) = 81$
$i.e. a^2 + b^2 + c^2 = 81 - 46 = 35$
Now$, a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (a + b + c)[(a^2 + b^2 + c^{2)} - (ab + bc + ca)]$
$= (9)[35 - 23]$
$= 9 \times 12$
$= 108$
Hence$, (a + b + c)^2 = 81$
So$, a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = 81$
$i.e. a^2 + b^2 + c^2 + 2(ab + bc + ca) = 81$
$i.e. a^2 + b^2 + c^2 + 2(23) = 81$
$i.e. a^2 + b^2 + c^2 = 81 - 46 = 35$
Now$, a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (a + b + c)[(a^2 + b^2 + c^{2)} - (ab + bc + ca)]$
$= (9)[35 - 23]$
$= 9 \times 12$
$= 108$
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