Find the HCF of the following pair of polynomials:
$x^{3} + y^{3}$ and $3 x^{2} - 3 y^{2}$

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Solution

We know that $H C F$ is the highest common factor.

Factorise $x^{3} + y^{3}$ as follows:

$x^{3} + y^{3} = (x + y) [x^{2} + y^{2} - (x \times y)] = (x + y) (x^{2} + y^{2} - x y)$
(using identity $a^{3} + b^{3} = (a + b) (a^{2} + b^{2} - a b)$ )

Now, factorise $3 x^{2} - 3 y^{2}$ as follows:

$3 x^{2} - 3 y^{2} = 3 (x^{2} - y^{2}) = 3 (x + y) (x - y)$ (using identity $a^{2} - b^{2} = (a + b) (a - b)$ )

Since the common factor between the polynomials $x^{3} + y^{3}$ and $3 x^{2} - 3 y^{2}$ is $(x + y)$ .

Hence, the $H C F$ is $(x + y)$ .

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x^{3} + 27

x^{2} - 9

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x^{3} - 3 x^{2} + 4 x - 12, x^{4} + x^{3} + 4 x^{2} + 4 x

If x + y + z = xyz. Then

$\frac{3 x - x^{3}}{1 - 3 x^{2}}$ + $\frac{3 y - y^{3}}{1 - 3 y^{2}}$ + $\frac{3 z - z^{3}}{1 - 3 z^{3}}$ =

x^{3} - 3 x^{2} + 4 x - 12, x^{4} + x^{3} + 4 x^{2} + 4 x

x + y + z = x y z

\frac{3 x - x^{3}}{1 - 3 x^{2}} + \frac{3 y - y^{3}}{1 - 3 y^{2}} + \frac{3 z - z^{3}}{1 - 3 z^{2}} = \frac{3 x - x^{3}}{1 - 3 x^{2}} \cdot \frac{3 y - y^{3}}{1 - 3 y^{2}} \cdot \frac{3 z - z^{3}}{1 - 3 z^{2}}

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