Basic Algebra Test Part 2

The given expression could be written as

$\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b} = \dfrac{ab+bc+ca}{ab+bc+ca}$

On Multiplying both the sides by $ab+bc+ca$ the expression becomes

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

$a^{2} +b^{2} + c^{2}- ab-bc-ca =-1(\dfrac{abc}{b+c}+\dfrac{abc}{a+c}+\dfrac{abc}{a+b} )$

We need to find

$\dfrac{a^{2}}{bc} + \dfrac{b^{2}}{ac} + \dfrac{c^{2}}{ab} = \dfrac{a^{3}+b^{3}+c^{3}}{abc}$

Apply the identitiy

$a^{3}+b^{3}+c^{3}=(a+b+c)(a^{2}+b^{2}+c^{2}- ab-bc-ca)+3abc$ in the numerator

Substitute the value of $a^{2}+b^{2}+c^{2}- ab-bc-ca$ to get

$\dfrac{-abc(\frac{a}{b+c}+1+\dfrac{b}{a+c}+1+\dfrac{c}{a+b}+1-3)}{abc}$

$=-(\dfrac{a}{b+c} + \dfrac{b}{a+c} + \dfrac{c}{a+b})$

=-1

$\large \frac{a}{b+c} + \frac{b} {a+c} + \frac{ c}{a+b} = 1$

Let us suppose that a=b

Therefore

$\implies \large 2.\frac{a}{a+c} +\frac{c}{2a}=1$

$\implies \large 2.\frac{1}{1+\frac{c}{a}} +\frac{1}{2}.\frac{c}{a}=1$

Assume $\frac{c}{a}=x$

$\implies \large 2.\frac{1}{1+x} +\frac{x}{2}=1$

$\implies x^2-x+2=0$ ........................................ $(1)$

REQUIRED = $\frac{ a^2}{b \times c} + \frac{ b^2}{a \times c} + \frac{ c^2} { a \times b}.$

using above assumption a=b

REQUIRED = $\large 2\frac{a}{c}+\frac{c^2}{a^2}$

$\large =x^2+\frac{2}{x}$

$\large =x-2+\frac{2}{x}$ because $x^2=x-2$ from (1)

$\large =\frac{x^2-2x+2}{x}$

$\large =\frac{x^2-x-x+2}{x}$ because $x^2-x=-2$ from (1)

$\large =\frac{-x}{x}=\boxed{-1}$ .

enjoy!

Note that if we replace $a, b, c$ by $ar, br, cr$ , the value of the expression, and the original condition don't change. ( $r$ is any complex number.)

Hence we let $abc=1$ . $a, b, c$ are nonzero otherwise the answer doesn't exist.

Multiplying the first equation by $(a+b)(b+c)(c+a)$ ,

$(a+b)(b+c)(c+a)=a (a+c)(a+b)+b (b+a)(b+c)+c(c+a)(c+b)=a^{3}+b^{3}+c^{3}+abc+(a+b)(b+c)(c+a)$ after expansion and factorisation.

Hence $a^{3}+b^{3}+c^{3}+abc=0$

$a^{3}+b^{3}+c^{3}=-1$

Dividing by $abc$ , we get the answer as $\frac{-1}{abc}=-1$ .

The given expression could be written as

(a+b+c) ( \frac {1} {b+c} + \frac {1} {a+c} + \frac {1} {a+b} ) = 4

Then we will get

(a+b+c)^3 = 3 (a+b+c)(ab+ac+bc) - 4abc

Consider

\frac {a^2} {bc} + \frac {b^2} {ac} + \frac {c^2} {ab} = k

Multiplying both the sides by abc

a^3 +b^3 + c^3 = kabc

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

-abc = kabc

So \boxed {k= -1}

yay :D