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Question Number 81857 by naka3546 last updated on 16/Feb/20

Commented by jagoll last updated on 16/Feb/20

$$zero=0$$

Commented by john santu last updated on 16/Feb/20

$$x^2+y^2=ab(a+b)+(a+b)(a^2-ab+b^2)$$$$x^2+y^2=(a+b)(a^2+b^2)\ast$$$$\Rightarrow (\frac{x}{a}-\frac{y}{b})(\frac{x}{b}-\frac{y}{a})=$$$$\frac{ab(x^2+y^2)-(a^2+b^2)xy}{{\left(ab\right)}^2}=$$$$\frac{ab(a+b)(a^2+b^2)-\left(ab\right)(a+b)(a^2+b^2)}{{\left(ab\right)}^2}=$$$$0.$$

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