Question

If \[a+b+c+d=4\] then find the value of \[\dfrac{1}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{1}{\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}+\dfrac{1}{\left( 1-c \right)\left( 1-d \right)\left( 1-a \right)}+\dfrac{1}{\left( 1-d \right)\left( 1-b \right)\left( 1-a \right)}\].

Answer 1

Hint: In this type of question we have to use the concept of LCM. In the given expression we observe the denominator of each term and then we multiply the numerator as well as the denominator of each of them by the bracket which is not present in the denominator. In this way we will get the same denominator for each term and then we can add the numerator. By simplifying the expression further we will reach the required value.

Complete step-by-step solution:
Now we have to find the value of \[\dfrac{1}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{1}{\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}+\dfrac{1}{\left( 1-c \right)\left( 1-d \right)\left( 1-a \right)}+\dfrac{1}{\left( 1-d \right)\left( 1-b \right)\left( 1-a \right)}\] and we have provided that \[a+b+c+d=4\].
Let us consider the given expression,
\[\Rightarrow \dfrac{1}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{1}{\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}+\dfrac{1}{\left( 1-c \right)\left( 1-d \right)\left( 1-a \right)}+\dfrac{1}{\left( 1-d \right)\left( 1-b \right)\left( 1-a \right)}\]
Here, we can observe that in first term \[\left( 1-d \right)\] is absent, in second term \[\left( 1-a \right)\], in third term \[\left( 1-b \right)\] and in fourth term \[\left( 1-c \right)\] are absent. So that we multiply each of the term by the corresponding terms respectively to obtain the same denominator which are as follows:
\[\begin{align}
  & \Rightarrow \text{First Term =}\dfrac{\left( 1-d \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)} \\
 & \Rightarrow \text{Second Term =}\dfrac{\left( 1-a \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)} \\
 & \Rightarrow \text{Third Term =}\dfrac{\left( 1-b \right)}{\left( 1-c \right)\left( 1-d \right)\left( 1-a \right)\left( 1-b \right)} \\
 & \Rightarrow \text{Fourth Term =}\dfrac{\left( 1-c \right)}{\left( 1-d \right)\left( 1-b \right)\left( 1-a \right)\left( 1-c \right)} \\
\end{align}\]
Hence, by adding all the terms we will get the above expression as,

\[\Rightarrow \dfrac{\left( 1-d \right)+\left( 1-a \right)+\left( 1-b \right)+\left( 1-c \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}\]
On simplification, we can write the above expression as
\[\begin{align}
  & \Rightarrow \dfrac{\left( 1+1+1+1 \right)-d-a-b-c}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)} \\
 & \Rightarrow \dfrac{4-\left( a+b+c+d \right)}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)} \\
\end{align}\]
But as we have given that the value of \[a+b+c+d=4\]
\[\Rightarrow \dfrac{4-4}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}\]
\[\Rightarrow 0\]
Hence, the value of the expression is equal to \[0\]
Thus, if \[a+b+c+d=4\] then the value of \[\dfrac{1}{\left( 1-a \right)\left( 1-b \right)\left( 1-c \right)}+\dfrac{1}{\left( 1-b \right)\left( 1-c \right)\left( 1-d \right)}+\dfrac{1}{\left( 1-c \right)\left( 1-d \right)\left( 1-a \right)}+\dfrac{1}{\left( 1-d \right)\left( 1-b \right)\left( 1-a \right)}\] is equal to \[0\].

Note: In this type of question students have to note that by multiplication to the numerator as well as to the denominator by an absent term from the denominator they can obtain the same denominator for each of the terms. Also students have to remember that zero divided by anything always results in a zero.