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Question

If $$\dfrac{a}{b}+\dfrac{b}{a}=-1$$ then find $$a^3-b^3$$.


Solution

$$\dfrac{a}{b}+\dfrac{b}{a}=-1$$
$$a^2+b^2=-ab$$
$$a^2+ab+b^2=0-----(1)$$
$$a^3-b^3=(a-b)(a^2+b^2+ab)$$
But from $$(1)$$
$$a^2+b^2+ab=0$$
$$a^3-b^3=(a-b)(0)=0$$

Mathematics

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