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Question Number 25948 by 272; 2; 2 last updated on 16/Dec/17

Factorise:  x^2 +y^2 −z^2 −2xy

$$\mathrm{Factorise}:\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} −{z}^{\mathrm{2}} −\mathrm{2}{xy} \\ $$

Commented by tawa tawa last updated on 16/Dec/17

x^2  + y^2  − z^2  − 2xy  rearrange  x^2  −2xy + y^2  − z^2   = x^2  − xy − xy + y^2  − z^2   = (x^2  − xy) − (xy + y^2 ) − z^2   = x(x − y) − y(x − y) − z^2   = (x − y)(x − y) − z^2   = (x − y)^2  − z^2   From difference of two square:  a^2  − b^2  = (a + b)(a − b)  where,  a = x − y  and  b = z  = (x − y)^2  − z^2   = (x − y + z)(x − y − z)

$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{z}^{\mathrm{2}} \:−\:\mathrm{2xy} \\ $$$$\mathrm{rearrange} \\ $$$$\mathrm{x}^{\mathrm{2}} \:−\mathrm{2xy}\:+\:\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{xy}\:−\:\mathrm{xy}\:+\:\mathrm{y}^{\mathrm{2}} \:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\left(\mathrm{x}^{\mathrm{2}} \:−\:\mathrm{xy}\right)\:−\:\left(\mathrm{xy}\:+\:\mathrm{y}^{\mathrm{2}} \right)\:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\mathrm{x}\left(\mathrm{x}\:−\:\mathrm{y}\right)\:−\:\mathrm{y}\left(\mathrm{x}\:−\:\mathrm{y}\right)\:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\left(\mathrm{x}\:−\:\mathrm{y}\right)\left(\mathrm{x}\:−\:\mathrm{y}\right)\:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\left(\mathrm{x}\:−\:\mathrm{y}\right)^{\mathrm{2}} \:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$\mathrm{From}\:\mathrm{difference}\:\mathrm{of}\:\mathrm{two}\:\mathrm{square}:\:\:\mathrm{a}^{\mathrm{2}} \:−\:\mathrm{b}^{\mathrm{2}} \:=\:\left(\mathrm{a}\:+\:\mathrm{b}\right)\left(\mathrm{a}\:−\:\mathrm{b}\right) \\ $$$$\mathrm{where},\:\:\mathrm{a}\:=\:\mathrm{x}\:−\:\mathrm{y}\:\:\mathrm{and}\:\:\mathrm{b}\:=\:\mathrm{z} \\ $$$$=\:\left(\mathrm{x}\:−\:\mathrm{y}\right)^{\mathrm{2}} \:−\:\mathrm{z}^{\mathrm{2}} \\ $$$$=\:\left(\mathrm{x}\:−\:\mathrm{y}\:+\:\mathrm{z}\right)\left(\mathrm{x}\:−\:\mathrm{y}\:−\:\mathrm{z}\right) \\ $$

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