Question

Given a+b+c=0 , abc ≠ 0 , a, b,c, are not equal to each other, find the value of a(\dfrac{1}{b}+\dfrac{1}{c})+b(\dfrac{1}{c}+\dfrac{1}{a})+c(\dfrac{1}{a}+\dfrac{1}{b})

Steven Zheng Steven Zheng posted 3 weeks ago


Answer

a(\dfrac{1}{b}+\dfrac{1}{c})+b(\dfrac{1}{c}+\dfrac{1}{a})+c(\dfrac{1}{a}+\dfrac{1}{b})

=\dfrac{a+c}{b}+\dfrac{a+b}{c}+\dfrac{b+c}{a}

=\dfrac{a^2c+ac^2+a^2b+ab^2+b^2c+bc^2}{abc}

=\dfrac{a^2(b+c)+b^2(a+c)+c^2(a+b)}{abc}

=\dfrac{(a+b+c)(a^2+b^2+c^2)-a^3-b^3-c^3}{abc}

=\dfrac{-a^3-b^3-c^3}{abc} (\because a+b+c=0)

Using the formula

a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)

We get,

\dfrac{-a^3-b^3-c^3}{abc}=-3

Steven Zheng Steven Zheng posted 3 weeks ago

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