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Algebra Level 4

If $\quad\dfrac{a}{b+c+d}+\dfrac{b}{a+c+d}+\dfrac{c}{a+b+d}+\dfrac{d}{a+b+c}=1,$ then what is the value of

$\dfrac{a^2}{b+c+d}+\dfrac{b^2}{a+c+d}+\dfrac{c^2}{a+b+d}+\dfrac{d^2}{a+b+c}?$

The answer is 0.

1 solution

Paola Ramírez
Jan 6, 2015

$(a+b+c+d)(\frac{a}{b+c+d}+\frac{b}{a+c+d}+\frac{c}{a+b+d}+\frac{d}{a+b+c})=1(a+b+c+d)$

$\frac{a^{2}+ab+ac+ad}{b+c+d}+\frac{ab+b^2+bc+bd}{a+c+d}+\frac{ac+bc+c^2+cd}{a+b+d}+\frac{ad+bd+cd+d^2}{a+b+c})=a+b+c+d$

$\frac{ab+ac+ad}{b+c+d}+\frac{a^2}{b+c+d}+\frac{ab+bc+bd}{a+c+d}+\frac{b^2}{b+c+d}+\frac{ac+bc+cd}{a+b+d}+\frac{c^2}{b+c+d}+\frac{ad+bd+cd}{a+b+c}+\frac{d^2}{b+c+d}=a+b+c+d$

$\frac{a^2}{b+c+d}+a+\frac{b^2}{a+c+d}+b+\frac{c^2}{a+b+d}+c+\frac{d^2}{a+b+c}+d=a+b+c+d$

$\boxed{\frac{a^2}{b+c+d}+\frac{b^2}{a+c+d}+\frac{c^2}{a+b+d}+\frac{d^2}{a+b+c}=0}$

as easy as eating a cake @paola Ramirez and nice solution

Mardokay Mosazghi - 6 years, 5 months ago

Nice way and a good question! cheers

Prachi Garg - 6 years, 5 months ago

That sure was a very nice solution.....enjoyed it........

Deep Chatterjee - 5 years, 10 months ago

wow ... a great solution, as easier than I expected! 😊👏🏼👏🏼👏🏼 Congratulations

Carlos Bravo - 5 years, 5 months ago

As easy as...

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