If $x^{2} + y^{2} = 1$ and $p = (3 x - 4 x^{3})^{2} + (3 y - 4 y^{3})^{2}$ then $P$ is equal to

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Solution

$x^{2} + y^{2} = 1 \Rightarrow y^{2} = 1 - x^{2}$
$P = {(3 x - 4 x^{3})}^{2} + {(3 y - 4 y^{3})}^{2}$
$= x^{2} {(3 - 4 x^{2})}^{2} + y^{2} {(3 - 4 y^{2})}^{2}$
$= x^{2} {(3 - 4 x^{2})}^{2} + (1 - x^{2}) {[3 - 4 (1 - x^{2})]}^{2}$
$= x^{2} {(3 - 4 x^{2})}^{2} + (1 - x^{2}) {(4 x^{2} - 1)}^{2}$
$= x^{2} (9 + 16 x^{4} - 24 x^{2}) + (1 - x^{2}) (16 x^{4} + 1 - 8 x^{2})$
$= 9 x^{2} + 16 x^{6} - 24 x^{4} + 16 x^{4} + 1 - 8 x^{2} - 16 x^{6} - x^{2} + 8 x^{4}$
$∴ P = 1$

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