If the values a + b + c = 8, ab + bc + ca = 17 and abc = 10 are given, then find the value of $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$? (a) 94 (b) 84 (c) 68 (d) 88
Answer
Verified
Hint: We start solving the problem by assuming the value of $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$ as ‘d’. We now multiply each term of $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$ present in brackets. After multiplication we substitute values given in the problem to get the required result.
Complete step by step answer: We have given values of a + b + c, ab + bc + ca and abc are 8, 17 and 10. We need to find the value of $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$. Let us first multiply all the given terms in $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$. Let us assume this value is ‘d’. So, $d=\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$. $\Rightarrow d=\left( 2.2+2.a+2.b+a.b \right).\left( 2+c \right)$. $\Rightarrow d=\left( 4+2a+2b+ab \right).\left( 2+c \right)$. $\Rightarrow d=\left( 4.2+4.c+2a.2+2a.c+2b.2+2b.c+ab.2+ab.c \right)$. $\Rightarrow d=8+4c+4a+2ac+4b+2bc+2ab+abc$. $\Rightarrow d=8+4a+4b+4c+2ab+2bc+2ca+abc$. $\Rightarrow d=8+4.\left( a+b+c \right)+2.\left( ab+bc+ca \right)+abc$ ---(1). We substitute the values a + b + c = 8, ab + bc + ca = 17 and abc = 10 in the equation (1). d = 8 + 4(8) + 2(17) + 10. d = 8 + 32 + 34 + 10. d = 84. ∴ The value of $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)=84$.
So, the correct answer is “Option B”.
Note: Alternatively, we can calculate the values of ‘a’, ‘b’ and ‘c’ to substitute in the given equation $\left( 2+a \right).\left( 2+b \right).\left( 2+c \right)$. We should not make any calculation mistakes while solving this problem to get the accurate result. Similarly, we can expect to find the polynomial having the roots ‘a’, ‘b’ and ‘c’.
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