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Factorisation by Regrouping Terms

Factorise : ...

Question

Factorise :
$(x^{2} - 2 x y + y^{2}) - z^{2}$

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Solution

$(x^{2} - 2 x y + y^{2}) - z^{2}$

$\Rightarrow [x (x - y) - y (x - y)] - z^{2}$

$\Rightarrow (x - y) (x - y) - z^{2}$

$\Rightarrow (x - y)^{2} - z^{2}$

$\Rightarrow [(x - y) + z] [(x - y) - z]$

$\Rightarrow (x - y + z) (x - y - z)$

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Similar questions

z^{2} - (x^{2} - 2 x y + y^{2})

Q. Factorisation of

z^{2} - (x^{2} + 2 x y + y^{2})

\frac{x^{2} + y^{2} - z^{2} + 2 x y}{x^{2} - y^{2} - z^{2} - 2 y z} .

Q. Factorise x(y+z)(y²+z²-x²)+y(z+x)(z²+x²-y²)+ z(x+y)(x²+y²-z²)

(x^{2} - 2 x y + y^{2}) - z^{2}

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