Question : If $x = 32.5$, $y = 34.6$ and $z = 30.9$, then the value of $x^3+y^3+z^3-3xyz$ is $0.98k$, where $k$ is equal to:
Option 1: 1033
Option 2: 933
Option 3: 1026
Option 4: 921
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Correct Answer: 1033
Solution :
Given that $x = 32.5$, $y = 34.6$ and $z = 30.9$
Use the identity,
$x^3+y^3+z^3-3xyz=\frac{1}{2}[x+y+z][(x-y)^2+(y-z)^2+(z-x)^2]$
$⇒0.98k=\frac{1}{2}[32.5+34.6+30.9][(32.5-34.6)^2+(34.6-30.9)^2+(30.9-32.5)^2]$
$⇒\frac{1}{2}[98][(-2.1)^2+(3.7)^2+(-1.6)^2]=0.98k$
$⇒49[4.41+13.69+2.56]=0.98k$
$⇒49[20.66]=0.98k$
$⇒k=1033$
Hence, the correct answer is 1033.
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