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Question Number 177928 by Spillover last updated on 11/Oct/22

$$\mathrm{Prove}\mathrm{that}$$$$\left[\frac{{\mathrm{x}}^3+{\mathrm{y}}^3}{{(\mathrm{x}-\mathrm{y})}^2+3\mathrm{y}}\right]\mathbf{÷}\left[\frac{{(\mathrm{x}+\mathrm{y})}^2-3\mathrm{xy}}{{\mathrm{x}}^3-{\mathrm{y}}^3}\right]\times \left[\frac{\mathrm{xy}}{{\mathrm{x}}^2-{\mathrm{y}}^2}\right]=\mathrm{xy}$$

Answered by Ar Brandon last updated on 11/Oct/22

$$\frac{x^3+y^3}{{(x-y)}^2+3yx}\cdot \frac{x^3-y^3}{{(x+y)}^2-3xy}\cdot \frac{xy}{x^2-y^2}$$$$=\frac{x^3+y^3}{x^2+xy+y^2}\cdot \frac{x^3-y^3}{x^2-xy+y^2}\cdot \frac{xy}{(x-y)(x+y)}$$$$=\frac{(x^3+y^3)(x-y)}{x^3-y^3}\cdot \frac{(x^3-y^3)(x+y)}{x^3+y^3}\cdot \frac{xy}{(x-y)(x+y)}$$$$=(x-y)(x+y)\cdot \frac{xy}{(x-y)(x+y)}=xy$$

Commented by Spillover last updated on 11/Oct/22

$$\mathrm{thank}\mathrm{you}$$

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