Question 803196
I do not like that problem. I think there is a typo.
I will solve the problem as typed, and the problem that was probably meant.
 
AS TYPED:
The problem as presented requires a lot of tedious work to get to 2 different ugly answers:
If {{{p-4y=2}}} & {{{py=8}}},
{{{y=8/p}}} & {{{p-4(8/p)=2}}}-->{{{p^2-32=2p}}}-->{{{p^2-2p-32=0}}}-->{{{p=1 +- sqrt(33)}}}
 
That's ugly, but it gets worse:
 
{{{p=1+sqrt(33)}}}-->{{{1+sqrt(33)-4y=2}}}-->{{{4y=-1+sqrt(33)}}}-->{{{y=(sqrt(33)-1)/4}}}
That gets us to {{{system(p^3=36sqrt(33)+100,y^3=(36sqrt(33)-100)/64)}}}-->{{{p^3-8y^3=31.5sqrt(33)+112.5=about293.45}}}
 
The other solution,
{{{p=1-sqrt(33)}}}-->{{{1-sqrt(33)-4y=2}}}-->{{{4y=-1-sqrt(33)}}}-->{{{y=-(sqrt(33)+1)/4}}}
That gets us to {{{system(p^3=100-36sqrt(33),y^3=-(36sqrt(33)+100)/64)}}}-->{{{p^3-8y^3=112.5-31.5sqrt(33)=about}}}{{{-68.45}}}
 
THE PROBLEM AS IT SHOULD BE:
{{{p-2y=4}}} & {{{py=8}}} allows for some clever maneuvering.
Since {{{p^3-8y^3=p^3-(2y)^3}}} ,
we could use some of those special products that the textbook represents as
{{{a^3-b^3=(a-b)(a^2+ab+b^2)}}} ,
{{{(a +- b)^2=a^2 +- 2ab+b^2}}} and
{{{(a +- b)^3=a^3 +- 3a^2b +3ab^2 +- b^3}}}
 
One way:
{{{p^3-8y^3=p^3-(2y)^3}}}
= {{{(p-2y)(p^2+p(2y)+(2y)^2)}}}
= {{{4(p^2+p(2y)+(2y)^2)}}}
= {{{4(p^2-2*p(2y)+(2y)^2+3p(2y))}}}
= {{{4((p-2y)^2+3p(2y))}}}
= {{{4(4^2+3p(2y))}}}
= {{{4(4^2+6py))}}}
= {{{4(4^2+6*8))=highlight(256)}}}
 
Another way to calculate {{{p^3-8y^3=p^3-(2y)^3}}}
{{{(p-2y)^3=p^3-3*p^2(2y)+3*p*(2y)^2-(2y)^3}}}
{{{4^3=p^3-3*p^2(2y)+3*p*(2y)^2-8y^3}}}
{{{64=p^3-8y^3-3*p^2(2y)+3*p*(2y)^2}}}
{{{64=p^3-8y^3-3*p(2y)(p-2y)}}}
{{{64=p^3-8y^3-6(py)(p-2y)}}}
{{{64=p^3-8y^3-6*8*4}}}
{{{64+6*8*4=p^3-8y^3}}}
{{{p^3-8y^3=highlight(256)}}}