We need to factorise,

[tex]\displaystyle\longrightarrow X=\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{c}+3[/tex]

That 3 can be written as,

• [tex]3=1+1+1=\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}[/tex]

So,

[tex]\displaystyle\longrightarrow X=\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{a}{a}+\dfrac{b}{b}+\dfrac{c}{c}[/tex]

Rearranging the terms like,

[tex]\displaystyle\longrightarrow X=\dfrac{a}{a}+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+\dfrac{b}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+\dfrac{c}{c}[/tex]

Taking [tex]a[/tex] common from first 3 terms, [tex]b[/tex] common from next 3 terms and [tex]c[/tex] common from last 3 terms,

[tex]\displaystyle\longrightarrow X=a\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+b\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)+c\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)[/tex]

Taking [tex]\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)[/tex] common from each term we get,

[tex]\displaystyle\longrightarrow\underline{\underline{X=\bigg(a+b+c\bigg)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)}}[/tex]

This is how we factorise the given expression.

Verified answer

Question :-

Resolve The fraction

[tex] \large \tt \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}+3 [/tex]

[tex]\rule{300pt}{0.1pt}[/tex]

[tex]\texttt{Answer: \( \tt (a b+b c+a c)(a+b+c) / a b c \)}[/tex]

[tex]\rule{300pt}{0.1pt}[/tex]

Step-by-step explanation:

[tex]\begin{gathered} \begin{array}{l} \\ \\ \\ \displaystyle\bf \qquad \frac{a}{b}+\frac{b}{a}+\frac{b}{c}+\frac{c}{b}+\frac{a}{c}+\frac{c}{a}+3 \\ \\ \\ \\ \displaystyle\bf \qquad=\Rightarrow \frac{\left(a^{2}+b^{2}\right) }{ a b}+ \frac{\left(b^{2}+c^{2}\right)}{ b c}+ \frac{\left(c^{2}+a^{2}\right) }{a c}+3 \\\\ \\ \\ \displaystyle\bf \qquad =\Rightarrow \frac{\left(c\left(a^{2}+b^{2}\right)+a\left(b^{2}+c^{2}\right)+b\left(a^{2}+c^{2}\right)\right)}{ a b c}+3 \\ \\ \\ \\ \displaystyle\bf \qquad=\Rightarrow \frac{\left(a^{2} c+b^{2} c+a b^{2}+a c^{2}+a^{2} b+c^{2} b\right) }{ a b c}+3 \\\\ \\ \\ \displaystyle\bf \qquad =\Rightarrow \frac{\left(a^{2} c+a c^{2}+b^{2} c+b c^{2}+a^{2} b+a b^{2}\right) }{a b c}+3 \\\\ \\ \\ \displaystyle\bf \qquad =\Rightarrow \frac{\left(a^{2} c+a c^{2}+b^{2} c+b c^{2}+a^{2} b+a b^{2}+3 a b c\right) }{a b c} \\\\ \\ \\ \displaystyle\bf \qquad =\Rightarrow \frac{\left(a^{2} c+a c^{2}+a b c+b^{2} c+b c^{2}+a b c+a^{2} b+a b^{2}+a b c\right) }{a b c} \\\\ \\ \\ \displaystyle\bf \qquad =\Rightarrow \frac{(a c(a+c+b)+b c(b+c+a)+a b(a+b+c)) }{a b c }\\\\ \\ \\ \fcolorbox{red}{pink}{ \boxed{ \color{blue} \displaystyle\bf \qquad =\Rightarrow \frac{(a b+b c+a c)(a+b+c) }{a b c}}} \end{array} \end{gathered}

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