Given #x+y=2# and #x^3+y^3=5#, what is #x^2+y^2#?

2 Answers

Answer:

It is #x^2+y^2=3#

Explanation:

From #x+y=2# we have that

#x+y=2=>(x+y)^2=2^2=>x^2+y^2+2xy=4#

From #x^3+y^3=5# we have that

#x^3+y^3=5=>(x+y)*(x^2+y^2-xy)=5=> x^2+y^2-xy=5/2#

So we know that

#x^2+y^2+2xy=4# (1)

and

#x^2+y^2-xy=5/2=>2*(x^2+y^2)-2xy=5# (2)

If you add (1) and (2) you get

#3*(x^2+y^2)=9=>x^2+y^2=3#

Dec 27, 2015

Answer:

#x^2 + y^2 = 3#

Explanation:

By applying the sum of cubes formula together with the given equations:

#5 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = 2(x^2 + y^2 - xy)#

#=> x^2 + y^2 - xy= 5/2#

#=>x^2 + y^2 = 5/2 + xy" "("*")#

Now, by cubing the first given equation, we get

#(x+y)^3 = 2^3#

#=>x^3 + 3x^2y + 3xy^2 + y^3 = 8#

#=> (x^3 + y^3) + 3xy(x + y) = 8#

Substituting in our known values for #x+y# and #x^3 + y^3#, we obtain

#=> 5 + 3xy*2 = 8#

#=> 6xy = 3#

#=> xy = 1/2#

Substituting this back into #("*")#:

#x^2 + y^2 = 5/2 + 1/2#

#:. x^2 + y^2 = 3#