Difference between revisions of "MIE 2016/Day 1/Problem 9"
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==Solution == | ==Solution == | ||
| + | We start by expanding <math>(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)</math>. | ||
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| + | As we are given <math>(x+y+z)</math> and <math>x^2+y^2+z^2</math>, we get <math>(xy+yz+xz)</math> is <math>12</math>. | ||
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| + | Next, we simplify the third case and obtain <math>4(xy+yz+xz)=xyz</math> | ||
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| + | As we know <math>(xy+yz+xz)</math> is <math>12</math>, we know <math>xyz</math> is <math>48</math> | ||
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| + | Next we expand <math>(x+y+z)^3=x^3+y^3+z^3+3((x+y+z)(xy+yz+xz)-xyz)</math> | ||
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| + | Rearranging the equation we arrive at <math>x^3+y^3+z^3=(x+y+z)^3-3((x+y+z)(xy+yz+xz)-xyz)</math> | ||
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| + | Substituting all the known values, we get <math>x^3+y^3+z^3=343-3((7)(12)-48) = 235</math>. <math>\boxed{\textbf{B}}</math>. | ||
==See Also== | ==See Also== | ||
Latest revision as of 22:40, 7 January 2018
Problem 9
Let 
, 
and 
be complex numbers that satisfies the following system:
Compute 
.
(a) 
(b) 
(c) 
(d) 
(e) 
Solution
We start by expanding 
.
As we are given 
and 
, we get 
is 
.
Next, we simplify the third case and obtain 
As we know is , we know 
is 
Next we expand 
Rearranging the equation we arrive at 
Substituting all the known values, we get 
. 
.
