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Factorisation Using Algebraic Identities

If p=x-y, q...

Question

If $p = x - y, q = y - z$ and $r = z - r$ , simplify $r^{2} - p^{2} + 2 p q - q^{2}$

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Solution

We know that $(a - b)^{2} = a^{2} + b^{2} - 2 a b$

It is given that $p = x - y$ , $q = y - z$ and $r = z - x$ and calculate $r^{2} - p^{2} + 2 p q - q^{2}$ using above identity as shown below:

$r^{2} - p^{2} + 2 p q - q^{2} = (z - x)^{2} - (x - y)^{2} + 2 (x - y) (y - z) - (y - z)^{2}$
$= z^{2} + x^{2} - 2 z x - (x^{2} + y^{2} - 2 x y) + 2 (x y - y^{2} - x z + y z) - (y^{2} + z^{2} - 2 y z)$
$= z^{2} + x^{2} - 2 x z - x^{2} - y^{2} + 2 x y + 2 x y - 2 y^{2} - 2 x z + 2 y z - y^{2} - z^{2} + 2 y z = 4 x y - 4 x z + 4 y z - 4 y^{2}$
$= 4 (x y - x z + y z - y^{2}) = 4 [x (y - z) - y (y - z)] = 4 (x - y) (y - z) = 4 p q$

Hence, $r^{2} - p^{2} + 2 p q - q^{2} = 4 p q$

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z^{\frac{1}{3}} = p + i q

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