Question : What will be the value of $x^{3}+y^{3}+z^{3}-3xyz$, when $x+y+z=9$ and $x^{2}+y^{2}+z^{2}=31?$
Option 1: 27
Option 2: 3
Option 3: 54
Option 4: 9
Correct Answer: 54
Solution :
Give: $x+y+z=9$ ......(1)
$x^{2}+y^{2}+z^{2}=31$ ......(2)
As $(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)$
Squaring (1) on both sides, we have,
$(x+y+z)^2 = 81$
⇒ $x^2 + y^2 + z^2 +2(xy+yz+zx) = 81$
⇒ $31+2(xy+yz+zx) = 81$
⇒ $xy+yz+zx = 25$
Now, $x^{3}+y^{3}+z^{3}-3xyz$
= $ (x+y+z)(x^2 + y^2 + z^2 - xy - yz- zx)$
= $(9)(31-25) = 9\times 6 = 54$
Hence, the correct answer is 54.
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