Question 1105564
<pre>
He's no more help than the back of the book. LOL
She didn't give the complex solutions.

{{{system((x+y)(x^2+y^2)=40,
(x-y)(x^2-y^2)=16)}}}

Notice that if we interchange x and y, the system of equations
simplifies to the same system, so if (x,y)=(p,q) is a solution,
then so is (x,y)=(q,p) 

Let y = kx

{{{system((x+kx)(x^2+k^2x^2)=40,
(x-kx)(x^2-k^2x^2)=16)}}}

{{{system(x(1+k)x^2(1+k^2)=40,
x(1-k)x^2(1-k^2)=16)}}}

{{{system(x^3(1+k)(1+k^2)=40,
x^3(1-k)(1-k^2)=16)}}}

Solve each for x³

{{{system(x^3=40/((1+k)(1+k^2)),
x^3=16/((1-k)(1-k^2)))}}}

Equate the expressions for x³

{{{40/(1+k)(1+k^2)=16/(1-k)(1-k^2)}}}

Divide both sides by 8

{{{5/(1+k)(1+k^2)=2/(1-k)(1-k^2)}}}

Cross-multiply:

{{{2(1+k)(1+k^2)=5(1-k)(1-k^2)}}}

Factor 1-k² as the difference of squares:

{{{2(1+k)(1+k^2)=5(1-k)(1-k)(1+k)}}}

Get 0 on the right

{{{2(1+k)(1+k^2)-5(1-k)(1-k)(1+k)=0}}}

Factor out common factor (1+k)

{{{(1+k)(2(1+k^2)^""-5(1-k)(1-k)^"")=0}}}

{{{(1+k)(2+2k^2-5(1-2k+k^2))=0}}}

{{{(1+k)(2+2k^2-5+10k-5k^2)=0}}}

{{{(1+k)(-3k^2 + 10k - 3)=0}}}

{{{(1+k)(-3k^2 + 10k - 3)=0}}}

Use the zero factor property:

1+k = 0;    -3k²+10k-3 = 0
  k = -1;    3k²-10k+3 = 0
           (3k-1)(k-3) = 0  
         3k-1 = 0;  k-3 = 0
           3k = 1;    k = 3
            k = 1/3

We substitute those three values for k in:

{{{x^3(1-k)(1-k^2)=16}}}

Substituting k=-1

{{{x^3(1-(-1)^"")(1-(-1)^2)=16}}}
{{{x^3(1+1)(1-1)=16}}}
{{{0=16}}}, not possible:

Substituting k=1/3

{{{x^3(1-(1/3)^"")(1-(1/3)^2)=16}}}
{{{x^3(2/3)(1-1/9)=16}}} 
{{{x^3(2/3)(8/9)=16}}}
{{{x^3(16/27)=16}}}
{{{x^3(16)=16*27}}}
{{{x^3=27}}}
{{{x^3-27=0}}}
{{{(x-3)(x^2+3x+9)=0 }}}}}}

Use zero-factor property:

x=3; x²+3x+9=0
     {{{x = (-3 +- sqrt( 3^2-4*9*1 ))/(2*1) }}}
     {{{x = (-3 +- sqrt(9-36))/2 }}}
     {{{x = (-3 +- sqrt(-27))/2 }}}
     {{{x = (-3 +- i*sqrt(27))/2 }}}
     {{{x = (-3 +- i*sqrt(9*3))/2 }}}
     {{{x = (-3 +- 3i*sqrt(3))/2 }}}

And since y = kx = (1/3)x

when x = 3, y = (1/3)(1) = 1

So one solution is (x,y) = (3,1), and by swapping x and y, we
have

 (x,y) = (1,3)

and when {{{x = (-3 +- 3i*sqrt(3))/2 }}},  {{{y = (1/3)((-3 +- 3i*sqrt(3))/2) = (-1 +- i*sqrt(3))/2 }}}

So two more solutions are

{{{matrix(1,5,
"(x,y)","",""="","",(matrix(1,3,(-3 + 3i*sqrt(3))/2,",",(-1 + i*sqrt(3))/2)))}}}

and

{{{matrix(1,5,
"(x,y)","",""="","",(matrix(1,3,(-3 - 3i*sqrt(3))/2,",",(-1 - i*sqrt(3))/2)))}}}

And two more solutions by swapping x and y are

{{{matrix(1,5,
"(x,y)","",""="","",(matrix(1,3,(-1 + i*sqrt(3))/2,",",(-3 + 3i*sqrt(3))/2)))}}}

and

{{{matrix(1,5,
"(x,y)","",""="","",(matrix(1,3,(-1 - i*sqrt(3))/2,",",(-3 - 3i*sqrt(3))/2)))}}}

Edwin</pre>