Question 167734
{{{system(x^3+y^3=1, x^2y+2xy^2+y^3=2)}}}
<pre><font size = 4 color = "indigo"><b>
Let {{{y=kx}}}

{{{system(x^3+(kx)^3=1, matrix(1,3,x^2(kx)+2x(kx)^2+(kx)^3,"=",2))}}}

{{{system(x^3+k^3x^3=1, matrix(1,3,kx^3+2x(k^2x^2)+k^3x^3,"=",2))}}}

{{{system(x^3+k^3x^3=1, matrix(1,3,kx^3+2k^2x^3+k^3x^3,"=",2))}}}

{{{system(x^3(1+k^3)=1, matrix(1,3,x^3(k+2k^2+k^3),"=",2))}}}

Now we can solve each for {{{x^3}}} as long as we require that
{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}

So we will take Case 1 as when they are both true:

Case 1:  {{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}

Then we can solve both for {{{x^3}}} without
dividing by 0:

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

Since the left sides are equal, so are the right 
sides:

{{{1/(1+k^3)=2/(k+2k^2+k^3)}}}

Cross multiply:

{{{2(1+k^3)=1(k+2k^2+k^3)}}}

{{{2+2k^3=k+2k^2+k^3}}}

Get 0 on the right and left side in descending
order:

{{{k^3-2k^2-k+2=0}}}

Factor k^2 out of the first two terms on the left,
and factor -1 out of the last two terms on the left:

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

Factor {{{k-2)}}} out of both terms on the left:

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

Factor the second parentheses on the left:

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

Using the zero-factor principle:

{{{matrix(2,5,k-2=0,"|",k-1=0,"|",k+1=0,       
                  k=2,"|", k=1,"|",k=-1)}}} 

Remember that the requirements for Case 1 were:

{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}

We investigate to see whether {{{k=2}}} satisfies
both of those requirements. Substituting,

{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}
{{{1+2^3<>0}}} and {{{2+2(2)^2+(2)^3<>0}}}
{{{1+8<>0}}} and {{{2+2(4)+8<>0}}}
{{{9<>0}}} and {{{2+8+8<>0}}}
{{{9<>0}}} and {{{18<>0}}}

So there are both satisfied, so we can use {{{k=2}}}

We now investigate to see whether {{{k=1}}} satisfies
both of those requirements. Substituting,

{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}
{{{1+1^3<>0}}} and {{{2+2(1)^2+(1)^3<>0}}}
{{{1+1<>0}}} and {{{2+2(1)+1<>0}}}
{{{2<>0}}} and {{{2+2+1<>0}}}
{{{2<>0}}} and {{{5<>0}}}

So these are both satisfied, so we can also use {{{k=1}}}

We now investigate to see whether {{{k=-1}}} satisfies
both of those requirements. Substituting,

{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}
{{{1+(-1)^3<>0}}} and {{{2+2(-1)^2+(-1)^3<>0}}}
{{{1-1<>0}}} and {{{2+2(1)-1<>0}}}
{{{0<>0}}} and {{{2+2-1<>0}}}
{{{0<>0}}} and {{{3<>0}}}

Since the first is not satisfied we cannot use {{{k=-1}}}

Now we use{{{k=1}}} and since {{{y=kx}}}, {{{y=x}}} in the original:

{{{system(x^3+y^3=1, x^2y+2xy^2+y^3=2)}}}
{{{system(x^3+x^3=1, x^2x+2xx^2+x^3=2)}}}
{{{system(2x^3=1, x^3+2x^3+x^3=2)}}}
{{{system(2x^3=1, 4x^3=2)}}}

solving both for {{{x^3}}}

{{{system(x^3=1/2,x^3=1/2)}}}

So these are the same, so we can

get the real solution by taking cube roots
of both sides:

{{{x=root(3,1/2)}}}

rationalizing:

{{{x=root(3,(1*4)/(2*4)) }}}

{{{x=root(3,4/8)}}}

{{{x=root(3,4)/2)}}}

And since in this case {{{y=x}}}

we have solution

{{{matrix(1,11,   "(", x, ",", y, ")", "=",  "(", root(3,4)/2, ",", root(3,4)/2, ")"  )}}} 

But that's not necessarily the only solution.

Now we use{{{k=2}}} and since {{{y=kx}}}, {{{y=2x}}} in the original:

{{{system(x^3+y^3=1, x^2y+2xy^2+y^3=2)}}}
{{{system(x^3+(2x)^3=1, x^2(2x)+2x(2x)^2+(2x)^3=2)}}}
{{{system(x^3+8x^3=1, 2x^3+2x(4x^2)+8x^3=2)}}}
{{{system(9x^3=1, 2x^3+8x^3+8x^3=2)}}}
{{{system(9x^3=1, 18x^3=2)}}}
{{{system(9x^3=1, 9x^3=1)}}}

solving both for {{{x^3}}}

{{{system(x^3=1/9,x^3=1/9)}}}

So these are the same, so we can
get a real solution by taking cube roots
of both sides:

{{{x=root(3,1/9)}}}

rationalizing:

{{{x=root(3,(1*3)/(9*3)) }}}

{{{x=root(3,3/27)}}}

{{{x=root(3,3)/3)}}}

And since in this case {{{y=2x}}}

we have solution

{{{matrix(1,11,   "(", x, ",", y, ")", "=",  "(", root(3,3)/3, ",", 2root(3,3)/3, ")"  )}}} 

So we have found two solutions.  But this
was only for Case 1, when 

{{{1+k^3<>0}}} and {{{k+2k^2+k^3<>0}}}

We must investigate the possibility of
solutions when one or both of these are
violated:

Case 2: {{{1+k^3=0}}}

This means that

{{{k^3=-1}}} or {{{k=-1}}}

Now we use{{{k=-1}}} and since {{{y=kx}}}, {{{y=-x}}} in the original:

{{{system(x^3+y^3=1, x^2y+2xy^2+y^3=2)}}}
{{{system(x^3+(-x)^3=1, x^2(-x)+2x(-x)^2+(-x)^3=2)}}}
{{{system(x^3-x^3=1, -x^3+2x^3-x^3=2)}}}
{{{system(0=1, 0=2)}}}

This is certainly not true, so Case 2
{{{1+k^3=0}}} is not possible since it produces
no solutions.

Case 3: {{{k+2k^2+k^3=0}}}

Factoring out k on the left:

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

Factoring the trinomial in parenheses:

{{{k(1+k)(1+k)=0}}}

Using the zero-factor principle:

{{{matrix(2,3,k=0,"|",1+k=0,       
                  "","|", k=-1)}}}

We have already seen that {{{k=-1}}} is not
possible, so we only need investigate {{{k=0}}}

Now we use{{{k=0}}} and since {{{y=0x}}}, {{{y=0}}} in the original:

{{{system(x^3+y^3=1, x^2y+2xy^2+y^3=2)}}}
{{{system(x^3+(0)^3=1, x^2(0)+2x(0)^2+(0)^3=2)}}}
{{{system(x^3=1, 0=2)}}}
{{{system(x=1, 0=2)}}}

This is certainly not true, so Case 3
{{{k=0}}} is not possible since it produces
no solutions.

So the only real solutions are:

{{{matrix(1,11,   "(", x, ",", y, ")", "=",  "(", root(3,4)/2, ",", root(3,4)/2, ")"  )}}}

and

{{{matrix(1,11,   "(", x, ",", y, ")", "=",  "(", root(3,3)/3, ",", 2root(3,3)/3, ")"  )}}} 

However, there are also some imaginary solutions.  I did not
find those. They would require finding the 2 imaginary
cube roots each of 4 and 3.  I only considered the real
cube roots of 4 and 3.  Post again if you want all the
imaginary solutions as well.

Edwin</pre>