Question
If $a + b + c = 9$ and $ab + bc + ca =23,$ then $a^3+ b^3+ c^3 - 3abc =$
✓
Answer
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$, a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (a + b + c)[(a^2 + b^2 + c^{2)} - (ab + bc + ca)]$
$= (9)[35 - 23]$
$= 9 \times 12$
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$, a^3 + b^3 + c^3 - 3abc$
$= (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$
$= (a + b + c)[(a^2 + b^2 + c^{2)} - (ab + bc + ca)]$
$= (9)[35 - 23]$
$= 9 \times 12$
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