Question

$\bf \: solve \: if \: you \: are \: genius. \\ \sf \:if \: (a + b + c) = abc \\ \\ \bf \: find : \frac{(1 - a^{2})(1 - b^{2})}{ab} + \frac{(1 - b^{2})(1 - c^{2})}{bc} + \frac{(1 - c^{2})(1 - a^{2})}{ca}. \\ \\ \sf \: full \: solution...$
​

Answer 1

$\underline{\green{\bf\:Given}} : \red{\sf(a + b + c) = abc} \\ \\ \underline{ \pink{\bf \:To\:Find}}: \blue{\sf \: \frac{(1 - a^{2})(1-{b}^{2})}{ab}+\frac{(1-b^{2})(1-{c}^{2})}{bc}+\frac{(1-c^{2})(1-a^{2})}{ca}} \\ \\ \large\overbrace{\underbrace{\pink{\boxed{\bf{\red{A}\green{n}s\blue{w}\orange{e}r:}}}}} \\ \\ \purple\longmapsto\tt(a + b + c) = abc \: \\ \sf \: dividing \: both \: sides \: by \: \red{b} \: we \: get,, \\ \\ \purple\longmapsto\tt \: \frac{a}{b} + 1 + \frac{c}{b} = ac \\ \\ \purple\longmapsto \boxed{\tt \: (\frac{a}{b} + \frac{c}{b}) = (ac - 1)} - - - \orange{\bf \: Equation(1)}\\ \\ \underline{\bf Similarly, \: dividing \: by \: a \: and \: c \: we \: Get}:- \\ \\\purple\longmapsto \boxed{\tt \: (\frac{b}{a}+\frac{c}{a}) = (bc - 1)} - - - \orange{\bf\:Equation(2)} \\ \\ \purple\longmapsto \boxed{\tt \: (\frac{a}{c}+\frac{b}{c})=(ab - 1)} - - - \orange{\bf \: Equation(3)}$

$\rule{200}{4}$

$\sf\dfrac{(1 - a^{2})(1-{b}^{2})}{ab}+\dfrac{(1-b^{2})(1-{c}^{2})}{bc}+\dfrac{(1-c^{2})(1-a^{2})}{ca} \\ \\ \underline{\bf \: Adding\:and\:Subtracting\: \red{ab\:,bc\:,ca}\:we\:get}:- \\ \\ \red\longmapsto\:\rm\dfrac{(1 - a^{2})(1-{b}^{2})}{ab}-ab+\dfrac{(1-b^{2})(1-{c}^{2})}{bc} - bc+\dfrac{(1-c^{2})(1-a^{2})}{ca} - ca \:+(ab + bc + ca) \\ \\ \red\longmapsto\:\rm\:\dfrac{(1-{b}^{2}-{a}^{2})}{ab}+\dfrac{(1-{b}^{2}-{c}^{2})}{bc}+\dfrac{(1-{c}^{2}-{a}^{2})}{ca} + (ab + bc + ca) \\ \\\red\longmapsto \rm \: ( \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}) - (\frac{{a}^{2}+{b}^{2}}{ab}+\frac{{b}^{2}+{c}^{2}}{bc}+\frac{{c}^{2}+{a}^{2}}{ca})+(ab + bc + ca) \\ \\ \red\longmapsto \pink{\boxed{\bf \: 1-(\frac{a}{b} +\frac{b}{a}+\frac{b}{c}+\frac{c}{b} +\frac{c}{a}+\frac{a}{c}) + (ab + bc + ca)}}$

$\rule{200}{4}$

$\underline\green{\textbf{Now,Putting value of Equation(1)}}\\ \underline\green{\textbf{(2) and(3) here we get}}:- \\ \\ \red\leadsto \sf \:1-(ac-1+bc-1+ab-1)+(ab + bc + ca) \\ \\ \red\leadsto \sf \: 1 + 1 + 1 + 1 \\ \\ \red\leadsto \: \bold{\boxed{\large{\boxed{\orange{\small{\boxed{\large{\red{\bold{4}}}}}}}}}} \: \underline{ \orange{\bf\:(Ans.)}}\: \boxed{\huge\bold{\red{\ddot{\smile}}}}$

$\rule{200}{4}$

Answer 2

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Question-}}}}}}$

$\bf \: solve \: if \: you \: are \: genius. \\ \sf \:if \: (a + b + c) = abc \\ \\ \bf \: find : \frac{(1 - a^{2})(1 - b^{2})}{ab} + \frac{(1 - b^{2})(1 - c^{2})}{bc} + \frac{(1 - c^{2})(1 - a^{2})}{ca}. \\ \\ \sf \: full \: solution...$

$\huge{\fbox{\fbox{\bigstar{\mathfrak{\red{Solution-}}}}}}$

(a+b+c) = abc

a/b + 1 + c/b = ac

____________________________________

(a/b + c/b) = (ac-1) ------equation 1

Similarly by dividing a and c we get,

(a/b)+(c/b) = ac -1 -----------equation2

And,

(a/c) + (b/c) = ab -1 --------equation3

____________________________________

=> (b/a+c/a) = (bc-1)

=> (a/c+b/c) = (ab-1)

____________________________________

adding, subtracting a and c,

=> (1-b²-a²)/ab + (1-c²-b²)/Bc +(1-a²-c²)/ca + [ab + Bc + ca]

=> (1/ab + 1/Bc + 1/ca )- (a²+b²/ab + b² +c²/Bc +c²+a²/ac) + (ab+Bc+ca)

=> 1 - (a/b + b/a + b/c + c/b +c/a + a/c) + (ab+bc+ca)

____________________________________

Now,putting values of equations

=> 1-(ac - 1 + bc-1 +ca-1 ) +(ab + Bc +ca)

=> 1+1+1+1

=>4

Hence,The answer is $4$