Question : If $(x+y+z)=12$, $xy+yz+zx=44$, and $xyz=48$, then what is the value of $x^{3}+y^{3}+z^{3}$?
Option 1: 104
Option 2: 144
Option 3: 196
Option 4: 288
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Correct Answer: 288
Solution :
Given: $(x+y+z)=12$
$xy+yz+zx=44$
$xyz=48$
We know that $x^{3}+y^{3}+z^{3}=(x+y+z)[(x+y+z)^{2}–3(xy+yz+zx)]+3xyz$
Putting the given values,
⇒ $x^{3}+y^{3}+z^{3}=12×[(12)^{2}–3(44)]+3×48$
⇒ $x^{3}+y^{3}+z^{3}=12×(144–132)+144$
⇒ $x^{3}+y^{3}+z^{3}=288$
Hence, the correct answer is 288.
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