Help me solve please

[Luke sammut]

x=\root(3)((p+q))+\root(3)((p-q))
and p^(2)-q^(2)=y^(2)
show that x^(3)-3yx-2p=0

 

[Deleted member 4993]

x=\root(3)((p+q))+\root(3)((p-q))
and p^(2)-q^(2)=y^(2)
show that x^(3)-3yx-2p=0

What do you need to solve?

 

[Denis]

x=\root(3)((p+q))+\root(3)((p-q))

WHAT is that? x = √3 * (p+q) + √3 * (p-q)?

If so, try it using p=5 and q=3 : what do you get for y ?

 

[Dr.Peterson]

x=\root(3)((p+q))+\root(3)((p-q))
and p^(2)-q^(2)=y^(2)
show that x^(3)-3yx-2p=0

I suspect that "\root(3)" is being taken to mean cube root.

Also, I suspect that the second equation is supposed to be p^2 - q^2 = y^3.

Then the problem is,

Given that x = cbrt(p+q) + cbrt(p-q) and y^3 = p^2 - q^2, show that x^3 - 3yx - 2p = 0.
​

I can do this, by cubing the expression for x, and replacing a piece of the result with x.

Have you tried this?

 

[Denis]

Given that x = cbrt(p+q) + cbrt(p-q) and y^3 = p^2 - q^2,
show that x^3 - 3yx - 2p = 0.
​

Yep...that works; nice catch Doc!