Question

if (x+a)(x+b)(x+c)=x^3+14x^2+59x+70, then find the value of
a) a+b+c
b) 1/a+1/b+1/c
c) a^2+b^2+c^2
d) a/bc+b/ac+c/ab​

Answer 1

Solution :-

$(x+a)(x+b)(x +c)=x^3+14x^2+59x+70$

$x^3+(a+b+c)x^2+(ab+bc+ca)x+abc=x^3+14x^2+59x+70$

Equating corresponding coefficients on both sides we get

$\boxed{a+b+c=14}.........(1)$

$ab+bc+ca=59.......(2)$

$abc=70.........(3)$

$\frac{(2)}{(3)}\implies\:\frac{ab+bc+ca}{abc}=\frac{59}{70}$

$\implies\:\frac{ab}{abc}+\frac{bc}{abc}+\frac{ca}{abc}=\frac{59}{70}$

$\implies\:\frac{1}{c}+\frac{1}{a}+\frac{1}{b}=\frac{59}{70}$

$\implies\:\boxed{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{59}{70}}$

we know that ,

$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$

$\implies\:14^2=a^2+b^2+c^2+2(59)$

$\implies\:196=a^2+b^2+c^2+118$

$\implies\:196-118=a^2+b^2+c^2$

$\implies\:\boxed{a^2+b^2+c^2=78}$..........(4)

$\frac{(4)}{(3)}\implies\:\frac{a^2+b^2+c^2}{abc}=\frac{78}{70}$

$\implies\:\frac{a^2}{abc}+\frac{b^2}{abc}+\frac{c^2}{abc}=\frac{39}{35}$

$\implies\:\boxed{\frac{a}{bc}+\frac{b}{ac}+\frac{c}{ab}=\frac{39}{35}}$

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