If we factor the left side of the second equation, we get the equation:

\((x+y)(x^2+xy+y^2)=162+x^2+y^2\)

Since the first equation gives x+y=10, the second equation now becomes:

\(10(x^2+xy+y^2)=162+x^2+y^2 \)

\(10x^2+10xy+10y^2= 162+x^2+y^2\)

\(9x^2+9y^2+10xy=162\)

Now substitute the first equation into the second

\(9(10-y)^2+9y^2+10(10-y)y=162\)

\(9(100-20y+y^2)+9y^2+100y-10y^2=162\)

\(900-180y+9y^2+9y^2+100y-10y^2=162\)

\(8y^2-80y+900=162\)

\(8y^2-80y+738=0\)

However, putting this into the quadratic formula, we find that the discriminant is negative, so therefore there are no real solutions.

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