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Question Number 49229 by munnabhai455111@gmail.com last updated on 04/Dec/18

$${\mathit{x}}^2-{\mathit{y}}^2$$

Commented by Abdo msup. last updated on 04/Dec/18

$$factorisation$$$$x^2-y^2=(x-y)(x+y)$$

Commented by munnabhai455111@gmail.com last updated on 04/Dec/18

thanks

Commented by hknkrc46 last updated on 04/Dec/18

$$=(x-y)(x+y)$$$$\mathit{o}\mathit{r}$$$$=(\sqrt{x^4-2x^2{y}^2+y^4}(\mid x\mid >\mid y\mid )$$$$\mathit{o}\mathit{r}$$$$=(\sqrt{x}-\sqrt{y})(\sqrt{x}+\sqrt{y})[{(\sqrt{x}+\sqrt{y})}^2-2\sqrt{xy}](x,y\nless 0)$$$$=(\sqrt{x}-\sqrt{y}){(\sqrt{x}+\sqrt{y})}^3-2\sqrt{xy}(\sqrt{x}-\sqrt{y})(x,y\nless 0)$$$$=(\sqrt{x}-\sqrt{y}){(\sqrt{x}+\sqrt{y})}^3-2x\sqrt{y}+2y\sqrt{x}(x,y\nless 0)$$$$\mathit{o}\mathit{r}$$$$=(x-k)(x+k)-(y-k)(y+k)(k\in \mathbb{R})$$$$\mathit{o}\mathit{r}$$$$=(x+\sqrt{y})(x-\sqrt{y})-[{(y-\frac{1}{2})}^2-\frac{1}{4}](y\nless 0)$$

Commented by Pk1167156@gmail.com last updated on 06/Dec/18

Very nice sir

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