Question 213425
{{{x^4+y^4=2}}}
and
{{{x^8+y^8=3}}}
then:
{{{x^16+y^16=8&1/2}}} 
how does it do so?

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<pre><font size = 4 color = "indigo"><b>
 
{{{x^4+y^4=2}}}
 
Square both sides:
 
{{{(x^4+y^4)^2=2^2}}}
 
{{{x^8+2x^4y^4+y^8=4}}}

{{{(x^8+y^8)+2x^4y^4=4}}}

but since {{{x^8+y^8=3}}}, we get

{{{3+2x^4y^4=4}}}  or subtracting 3 
from both sides

{{{2x^4y^4=1}}}

divide both sides by 2:

{{{x^4y^4=1/2}}}

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Now we turn to 

{{{x^8+y^8=3}}}
 
Square both sides:
 
{{{(x^8+y^8)^2=3^2}}}
 
{{{x^16+2x^8y^8+y^16=9}}}

{{{(x^16+y^16)+2x^8y^8=9}}}

Isolate {{{x^16+y^16}}}

{{{x^16+y^16=9-2x^8y^8}}}

Rewrite {{{x^8y^8}}} as{{{(x^4y^4)^2}}}

{{{x^16+y^16=9-2(x^4y^4)^2}}}

but since {{{x^4y^4=1/2}}}, we get

{{{x^16+y^16=9-2(1/2)^2}}}

{{{x^16+y^16=9-2(1/4)}}}

{{{x^16+y^16=9-1/2}}}

{{{x^16+y^16=8&1/2}}}

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