Question

Advanced: a,b,c are positive numbers such that a+b+ab= 8, b+c+bc=15 and c+a+ca = 35 What is the value of a+b+c+abc?
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Answer 1

$\huge\text{36}$

$\large\underline{\large\underline{\text{How to solve?}}}$

If we add 1, the factors can be separated. Since we know that all the three factors recur twice, we also know that each exponent will be 2.

To reduce the equation, we take the positive square root on both sides.

$\large\underline{\large\underline{\text{Explanation}}}$

We have to solve,

$\cdots \longrightarrow \begin{cases} & a+b+ab=8 \\ & b+c+bc=15 \\ & c+a+ca=35. \end{cases}$

It is equivalent to,

$\cdots \longrightarrow \begin{cases} & 1+a+b+ab=9 \\ & 1+b+c+bc=16 \\ & 1+c+a+ca=36. \end{cases}$

Let equation 1 be,

$\cdots \longrightarrow (a+1)(b+1)=9.\ \cdots (1)$

Let equation 2 be,

$\cdots \longrightarrow (b+1)(c+1)=16.\ \cdots (2)$

Let equation 3 be,

$\cdots \longrightarrow (c+1)(a+1)=36.\ \cdots (3)$

By $(1) \times (2) \times (3),$

$\cdots \longrightarrow \{(a+1)(b+1)(c+1)\}^{2}=(3 \cdot 4 \cdot 6)^{2}$

Let's recall that three numbers are positive, so,

$\cdots \longrightarrow (a+1)(b+1)(c+1)=3 \cdot 4 \cdot 6.\ \cdots (4)$

By $(4) \div (2),$

$\cdots \longrightarrow a+1= \dfrac{9}{2} \iff a=\dfrac{7}{2} .$

By $(4) \div (3),$

$\cdots \longrightarrow b+1=2 \iff b=1.$

By $(4) \div (1),$

$\cdots \longrightarrow c+1=8 \iff c=7.$

Hence,

$\cdots \longrightarrow a+b+c+abc=36.$

Answer 2

$\text{Given : a, b, c are positive numbers such that} \\ \frak{a+b+ab= 8, b+c+bc=15 \: and \: c+a+ca = 35}. \\ \text{What is the value of} \: \frak{ a+b+c+abc?}$

$\text{So, as per the let's follow the steps to get the required answer}$

$\dashrightarrow \frak{a + b + ab = 8} \\ \\ \dashrightarrow \frak{a(1 + b) + b = 8} \\ \\ \dashrightarrow \frak{a(1 + b) = 8 - b} \\ \\ \dashrightarrow \green{\frak{a = \frac{8 - b}{1 + b}}}$

$\text{Now, getting the second one} \\ \hookrightarrow \frak{b + c + bc = 15} \\ \\ \hookrightarrow \frak{b + c(1 + b) = 15} \\ \\ \hookrightarrow \frak{c(1 + b) = 15 - b} \\ \\ \hookrightarrow \red{\frak{c = \frac{15 - b}{1 + b}}}$

$\text{Finally getting the last one out} \\ \looparrowright \frak{c + a + ca = 35} \\ \\ \looparrowright \frak{ \frac{15 - b}{1 + b} + \frac{8 - b}{1 + b} + (\frac{15 - b}{1 + b})( \frac{8 - b}{1 + b}) = 35} \\ \\ \looparrowright \frak{ \frac{15 - b + 8 - b}{1 + b} + \frac{120 - 15b - 8b + {b}^{2} }{(1 + b)^{2} } = 35} \\ \\ \looparrowright \frak{(23 - 2b)(1 + b) + 120 - 23b + b ^{2} = 35 {(1 + b)}^{2} } \\ \\ \looparrowright \frak{36 {b}^{2} + 72b - 108 = 0 } \\ \qquad \rm{now \: following \: quadratic \: equations \: we \: get} \\ \\ \looparrowright \frak{(b + 3)(b - 1) = 0} \\ \\ \blue{\frak{b = 1}} \frak{(as \: all \: are \: positive)}$

$\sf{As \: we \: got \: b's \: value \: so \: let's \: put \: them \: in \: the \: first \: 2 \: cases \: to \: get \: a \: and \: c }$

$\text{value of a} \\ \quad : \implies \frak{ \frac{8 - b}{1 + b} } \\ \\: \implies \frak{ \frac{8 - 1}{1 + 1} } \\ \\: \implies \frak{ \frac{7}{2} } \\ \\ \text{value of c} \\ \quad : \implies \frak{ \frac{15- b}{1 + b} } \\ \\: \implies \frak{ \frac{15- 1}{1 + 1} } \\ \\ : \implies \frak{ \frac{14}{2} } \\ \\: \implies \frak{ 7 }$

$\text{ Getting the value of } \: \frak{a + b + c + abc } \\ \quad : \implies \frak{ \frac{7}{2} + 1 + 7 + \frac{7}{2} \times 1 \times 7 } \\ \\ \quad : \implies \frak{ \frac{7 + 2 + 14}{2} + \frac{49}{2} } \\ \\ \quad : \implies \frak{ \frac{23}{2} + \frac{49}{2}} \\ \\ \quad : \implies \frak{ \frac{72}{2}} \\ \\ \qquad \star \quad \boxed{ \pink{\frak{ 36 }}} \\ \\ \therefore\underline{\green{\sf{36 \: is \: the \: required \: answer}}}$