Question

If a/b +b/a =-1 then a^3-b^3 plz solve fastttt​

Answer 1

Answer:

$\sf{The \ value \ of \ a^{3}-b^{3} \ is \ 0.}$

Given:

$\sf{\dfrac{a}{b}+\dfrac{b}{a}=-1}$

To find:

$\sf{The \ value \ of \ a^{3}-b^{3}.}$

Solution:

$\sf{\leadsto{\dfrac{a}{b}+\dfrac{b}{a}=-1}}$

$\sf{\leadsto{\dfrac{a^{2}+b^{2}}{ab}=-1}}$

$\sf{\leadsto{a^{2}+b^{2}=-ab}}$

$\sf{Adding \ ab \ on \ both \ sides, \ we \ get}$

$\sf{\leadsto{a^{2}+b^{2}+ab=0...(1)}}$

$\sf{But,}$

$\sf{a^{3}-b^{3}=(a-b)(a^{2}+b^{2}+ab)}$

$\sf{From \ equation (1)}$

$\sf{\leadsto{a^{3}-b^{3}=(a-b)\times0}}$

$\sf{\leadsto{\therefore{a^{3}-b^{3}=0}}}$

$\sf\purple{\tt{\therefore{The \ value \ of \ a^{3}-b^{3} \ is \ 0.}}}$

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Extra information:

$\sf{a^{3}+b^{3}=(a+b)(a^{2}-ab+b^{2})}$

$\sf{a^{3}+b^{3}=(a+b)^{3}-3ab(a+b)}$

$\sf{(a+b)^{2}=(a-b)^{2}+4ab}$

$\sf{(a-b)^{2}=(a+b)^{2}-4ab}$