Question : If $x = 5.51, y = 5.52$, and $z = 5.57$, then what is the value of $x^3+y^3+z^3-3xyz$?
Option 1: 5.146
Option 2: 51.46
Option 3: 0.05146
Option 4: 0.5146
Correct Answer: 0.05146
Solution :
Given: The values of $x = 5.51, y = 5.52$, and $z = 5.57$
Use the algebraic identity, $x^3+y^3+z^3-3xyz=\frac{1}{2}(x+y+z)[(x-y)^2+(y-z)^2+(z-x)^2]$
So, $x^3+y^3+z^3-3xyz$
$=\frac{1}{2}(5.51+5.52+5.57)[(5.51-5.52)^2+(5.52-5.57)^2+(5.57-5.51)^2]$
$=\frac{1}{2}\times 16.6\times [(–0.01)^2+(–0.05)^2+(0.06)^2]$
$=8.3\times [0.0001+0.0025+0.0036]$
$=8.3\times 0.0062=0.05146$
Hence, the correct answer is 0.05146.
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