#### Answer

We show that
${\left( {{x^2} + {y^2} - x} \right)^2} = {\left( {1 + \cos \theta } \right)^2}$
${x^2} + {y^2} = {\left( {1 + \cos \theta } \right)^2}$,
thus,
${\left( {{x^2} + {y^2} - x} \right)^2} = {x^2} + {y^2}$.

#### Work Step by Step

The cardioid of Exercise 31 is given in polar coordinates by $r=1+\cos \theta$.
So, the rectangular coordinates of a point on the curve corresponding to $\theta$ is given by
$\left( {x,y} \right) = \left( {r\cos \theta ,r\sin \theta } \right)$,
$\left( {x,y} \right) = \left( {1 + \cos \theta } \right)\left( {\cos \theta ,\sin \theta } \right)$.
We have
$x = \left( {1 + \cos \theta } \right)\cos \theta $ ${\ \ \ }$ and ${\ \ \ }$ $y = \left( {1 + \cos \theta } \right)\sin \theta $.
Substituting $x$ and $y$ in the equation ${\left( {{x^2} + {y^2} - x} \right)^2}$ gives
${\left( {{x^2} + {y^2} - x} \right)^2} = {\left( {{{\left( {1 + \cos \theta } \right)}^2}{{\cos }^2}\theta + {{\left( {1 + \cos \theta } \right)}^2}{{\sin }^2}\theta - \left( {1 + \cos \theta } \right)\cos \theta } \right)^2}$
${\left( {{x^2} + {y^2} - x} \right)^2} = {\left( {{{\left( {1 + \cos \theta } \right)}^2} - \left( {1 + \cos \theta } \right)\cos \theta } \right)^2}$
${\left( {{x^2} + {y^2} - x} \right)^2} = {\left( {1 + 2\cos \theta + {{\cos }^2}\theta - \cos \theta - {{\cos }^2}\theta } \right)^2}$
${\left( {{x^2} + {y^2} - x} \right)^2} = {\left( {1 + \cos \theta } \right)^2}$
Since
${x^2} + {y^2} = {\left( {1 + \cos \theta } \right)^2}{\cos ^2}\theta + {\left( {1 + \cos \theta } \right)^2}{\sin ^2}\theta = {\left( {1 + \cos \theta } \right)^2}$,
therefore
${\left( {{x^2} + {y^2} - x} \right)^2} = {x^2} + {y^2}$.