Question 1095680
<font color="black" face="times" size="3">Given:
x+y = 4
xy = 2


We'll call the equations shown above to be <font color=blue>equation (1)</font> and <font color=blue>equation (2)</font> (in that order)


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Multiply both sides of <font color=blue>equation (2)</font> by 2


xy = 2


2*xy = 2*2


2xy = 4


We'll refer to this as <font color=blue>equation (3)</font>.


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Square both sides of <font color=blue>equation (1)</font>


x+y = 4


(x+y)^2 = 4^2


x^2+2xy+y^2 = 16


x^2+4+y^2 = 16 ....note how the "2xy" term has been replaced with 4. This is valid due to <font color=blue>equation (3)</font>


x^2+4+y^2-4 = 12-4 ... subtract 4 from both sides


x^2+y^2 = 12


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We will refer to x^2+y^2 = 12 as <font color=blue>equation (4)</font>.


Cube both sides of <font color=blue>equation (4)</font> to get the following:


x^2+y^2 = 12


(x^2+y^2)^3 = 12^3


(x^2)^3+3(x^2)^2(y^2)+3(x^2)(y^2)^2+(y^2)^3 = 1728


x^6+3x^4y^2+3x^2y^4+y^6 = 1728


(x^6+y^6)+3x^4y^2+3x^2y^4 = 1728


(x^6+y^6)+3x^2y^2(x^2+y^2) = 1728


(x^6+y^6)+3(xy)^2(x^2+y^2) = 1728


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Now make substitutions. We'll use <font color=blue>equation (2)</font> and <font color=blue>equation (4)</font>


(x^6+y^6)+3(xy)^2(x^2+y^2) = 1728


(x^6+y^6)+3(2)^2(x^2+y^2) = 1728 ... Use <font color=blue>equation (2)</font> to make the first substitution. Replace xy with 2.


(x^6+y^6)+3(2)^2(12) = 1728 ... Use <font color=blue>equation (4)</font> to make the next substitution. Replace x^2+y^2 with 12


(x^6+y^6)+144 = 1728


(x^6+y^6)+144-144 = 1728-144 ... Subtract 144 from both sides


x^6+y^6 = <font color=red>1584</font>


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Final Answer: <font color=red>1584</font>


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