Given $x + y = 2$ $x+y=2$ and $x^{3} + y^{3} = 5$ $x^3+y^3=5$ , what is $x^{2} + y^{2}$ $x^2+y^2$ ?

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Given $x + y = 2$ $x+y=2$ and $x^{3} + y^{3} = 5$ $x^3+y^3=5$ , what is $x^{2} + y^{2}$ $x^2+y^2$ ?

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Answer 1 · 2015-12-27T01:05:46

It is $x^{2} + y^{2} = 3$ $x^2+y^2=3$

Explanation:

From $x + y = 2$ $x+y=2$ we have that

$x + y = 2 \Rightarrow {(x + y)}^{2} = 2^{2} \Rightarrow x^{2} + y^{2} + 2 x y = 4$ $x+y=2=>(x+y)^2=2^2=>x^2+y^2+2xy=4$

From $x^{3} + y^{3} = 5$ $x^3+y^3=5$ we have that

$x^{3} + y^{3} = 5 \Rightarrow (x + y) \cdot (x^{2} + y^{2} - x y) = 5 \Rightarrow x^{2} + y^{2} - x y = \frac{5}{2}$ $x^3+y^3=5=>(x+y)*(x^2+y^2-xy)=5=> x^2+y^2-xy=5/2$

So we know that

$x^{2} + y^{2} + 2 x y = 4$ $x^2+y^2+2xy=4$ (1)

and

$x^{2} + y^{2} - x y = \frac{5}{2} \Rightarrow 2 \cdot (x^{2} + y^{2}) - 2 x y = 5$ $x^2+y^2-xy=5/2=>2*(x^2+y^2)-2xy=5$ (2)

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

$3 \cdot (x^{2} + y^{2}) = 9 \Rightarrow x^{2} + y^{2} = 3$ $3*(x^2+y^2)=9=>x^2+y^2=3$

Answer 2 · 2015-12-27T01:15:55

$x^{2} + y^{2} = 3$ $x^2 + y^2 = 3$

Explanation:

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

$5 = x^{3} + y^{3} = (x + y) (x^{2} - x y + y^{2}) = 2 (x^{2} + y^{2} - x y)$ $5 = x^3 + y^3 = (x+y)(x^2-xy+y^2) = 2(x^2 + y^2 - xy)$

$\Rightarrow x^{2} + y^{2} - x y = \frac{5}{2}$ $=> x^2 + y^2 - xy= 5/2$

$\Rightarrow x^{2} + y^{2} = \frac{5}{2} + x y (*)$ $=>x^2 + y^2 = 5/2 + xy" "("*")$

Now, by cubing the first given equation, we get

${(x + y)}^{3} = 2^{3}$ $(x+y)^3 = 2^3$

$\Rightarrow x^{3} + 3 x^{2} y + 3 x y^{2} + y^{3} = 8$ $=>x^3 + 3x^2y + 3xy^2 + y^3 = 8$

$\Rightarrow (x^{3} + y^{3}) + 3 x y (x + y) = 8$ $=> (x^3 + y^3) + 3xy(x + y) = 8$

Substituting in our known values for $x + y$ $x+y$ and $x^{3} + y^{3}$ $x^3 + y^3$ , we obtain

$\Rightarrow 5 + 3 x y \cdot 2 = 8$ $=> 5 + 3xy*2 = 8$

$\Rightarrow 6 x y = 3$ $=> 6xy = 3$

$\Rightarrow x y = \frac{1}{2}$ $=> xy = 1/2$

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

$x^{2} + y^{2} = \frac{5}{2} + \frac{1}{2}$ $x^2 + y^2 = 5/2 + 1/2$

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

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Given $x + y = 2$ $x+y=2$ and $x^{3} + y^{3} = 5$ $x^3+y^3=5$ , what is $x^{2} + y^{2}$ $x^2+y^2$ ?

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Given $x + y = 2$ $x+y=2$ and $x^{3} + y^{3} = 5$ $x^3+y^3=5$ , what is $x^{2} + y^{2}$ $x^2+y^2$ ?