Question

$\bf \: solve \: \: this \: \\ \\ \sf \:if \: (a + b + c) = abc \\ \\ \bf \: find : \frac{(1 - a^{2})(1 - b^{2})}{ab} + \frac{(1 - b... ^{2})(1 - c^{2})}{bc} + \frac{(1 - c^{2})(1 - a^{2})}{ca}.$
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Answer 1

$\large{\text{\underline{Short note:-}}}$

For this question to be defined, we need $abc\neq 0$.

$\large{\text{\underline{Solution:-}}}$

From transposition, we get the following equation.

$\hookrightarrow abc-a-b-c=0$

And now,

$\hookrightarrow \dfrac{abc-a-b-c}{a}=0$

$\hookrightarrow \dfrac{abc-a-b-c}{b}=0$

$\hookrightarrow \dfrac{abc-a-b-c}{c}=0$

Which shows,

$\hookrightarrow bc-1-\dfrac{b}{a}-\dfrac{c}{a}=0$

$\hookrightarrow ca-\dfrac{a}{b}-1-\dfrac{c}{b}=0$

$\hookrightarrow ab-\dfrac{a}{c}-\dfrac{b}{c}-1=0$

Adding all and by transposition,

$\hookrightarrow ab+bc+ca-(\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c})=3$

The three individual terms of the given expression are,

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

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

③$\dfrac{(1-c^{2})(1-a^{2})}{ca} =(\dfrac{1}{c}-c)(\dfrac{1}{a}-a)=ca-\dfrac{c}{a}-\dfrac{a}{c}+\dfrac{1}{ca}$

Lastly,

$\hookrightarrow \text{(Given expression)}=ab+bc+ca-(\dfrac{b}{a}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{c})+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}$

$\hookrightarrow \text{(Given expression)}=3+\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}$

$\hookrightarrow \text{(Given expression)}=3+\dfrac{a+b+c}{abc}$

$\hookrightarrow \text{(Given expression)}=4$

This is indeed a cool question! The answer to the question is 4. This is the end of the required answer.

$\large{\text{\underline{Exercise to the readers:-}}}$

This is an example of adding all the equations.

Find the value of $17x+9y+4z$, if $\begin{cases} &8x+3y+z=4 \\ &3x+y+z=3 \end{cases}$.

$\large{\text{\underline{Exercise solution:-}}}$

We cannot solve the equations, and it is the point of the question. It is two system equations in three variables, which lack the information.

Multiply both of the equations by $\alpha,\beta$ respectively.

$\hookrightarrow \begin{cases} &8\alpha x+3\alpha y+\alpha z=4\alpha \\ &3\beta x+\beta y+\beta z=3\beta \end{cases}$

Adding both equations,

$\hookrightarrow (8\alpha+3\beta)x+(3\alpha+\beta)y+(\alpha+\beta)z=4\alpha+3\beta$

Comparing the coefficients,

$\hookrightarrow \begin{cases} & 8\alpha+3\beta= 17\\ & 3\alpha+\beta= 9\\ & \alpha+\beta= 4\end{cases}\implies \alpha=1,\beta=3$

Hence,

$\hookrightarrow (8x+3y+z)+3(3x+y+z)=4+9=13$

13 is the required answer for this part.

Thank you.