Question

# 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

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.

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’.