Question

If a+b+c=22 and ab+bc+ca=91abc, then what is the value of $\dfrac{{a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right)}}{{abc}}$.
$
  {\text{A}}{\text{. 2002}} \\
  {\text{B}}{\text{. 484}} \\
  {\text{C}}{\text{. 1999}} \\
  {\text{D}}{\text{. 968}} \\
6 $

Answer 1

Hint- Here, we will proceed by simply multiplying the given two equations and then the obtained equation is somehow simplified in terms of the expression whose value is required such that the LHS is the expression and the RHS is the value corresponding to that expression.

Complete step-by-step answer:
Given, $a + b + c = 22{\text{ }} \to {\text{(1)}}$ and $ab + bc + ca = 91abc{\text{ }} \to {\text{(2)}}$
Two equations are multiplied with each other by simply multiplying the LHS of both the equations and the RHS of both the equations and then equating them.
By multiplying the given two equations (1) and (2), we get
$\left( {a + b + c} \right)\left( {ab + bc + ca} \right) = 22\left( {91abc{\text{ }}} \right)$
By simplifying the above equation, we get
$
   \Rightarrow {a^2}b + abc + c{a^2} + a{b^2} + {b^2}c + abc + abc + b{c^2} + {c^2}a = 2002abc \\
   \Rightarrow {a^2}b + c{a^2} + a{b^2} + {b^2}c + b{c^2} + {c^2}a + 3abc = 2002abc \\
 $
By taking the term 3abc on the RHS, the above equation becomes
$
   \Rightarrow {a^2}b + c{a^2} + a{b^2} + {b^2}c + b{c^2} + {c^2}a = 2002abc - 3abc \\
   \Rightarrow {a^2}b + c{a^2} + a{b^2} + {b^2}c + b{c^2} + {c^2}a = 1999abc{\text{ }} \to {\text{(3)}} \\
 $
Since, we have to find the value of the expression $\dfrac{{a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right)}}{{abc}}$ so the equation (3) can be rearranged as
$ \Rightarrow a{b^2} + {c^2}a + b{c^2} + {a^2}b + c{a^2} + {b^2}c = 1999abc$
Now taking a, b and c common from the first two terms, the next two terms and the last two terms on the LHS of the above equation, we get
$ \Rightarrow a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right) = 1999abc$
By shifting the term abc from the RHS to the LHS of the above equation, we get
$ \Rightarrow \dfrac{{a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right)}}{{abc}} = 1999$
So, the value of the expression $\dfrac{{a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right)}}{{abc}}$ is 1999.
Hence, option C is correct.

Note- In this particular problem, by observing carefully the expression whose value is required i.e., $\dfrac{{a\left( {{b^2} + {c^2}} \right) + b\left( {{c^2} + {a^2}} \right) + c\left( {{a^2} + {b^2}} \right)}}{{abc}}$ we can see that the two given equation should be multiplied together and by rearranging the obtained equation in order to take some terms common, we can get the required value of the expression.