Question

Determine whether the expression $\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$ can be expressed as ${{\left( 1-\sum{xy} \right)}^{2}}+{{\left( \sum{x-xyz} \right)}^{2}}$. If yes, enter the answer as 1. Else enter the answer as 0.

Answer 1

Hint: Individually evaluate and simplify both the terms in the question and then check whether they are the same or not. First, simplify the first term to arrive at a simpler expression. Then simplify the second term to arrive at the same expression. Do not open all the brackets at once as it will unnecessarily increase the calculation. Open them one by one and sequentially.
Complete step-by-step answer:
We have to check whether $\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$ can be expressed as ${{\left( 1-\sum{xy} \right)}^{2}}+{{\left( \sum{x-xyz} \right)}^{2}}$ .
Let us first simplify the first term which is $\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$.
We will open the first two brackets initially which gives us:
\[\left[ 1\times \left( 1+{{y}^{2}} \right)+{{x}^{2}}\times \left( 1+{{y}^{2}} \right) \right]\left( 1+{{z}^{2}} \right)\]
$\left( 1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$
Now, we will open the remaining brackets:
\[1\times \left( 1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}} \right)+{{z}^{2}}\times \left( 1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}} \right)\]
\[1+{{x}^{2}}+{{y}^{2}}+{{x}^{2}}{{y}^{2}}+{{z}^{2}}+{{x}^{2}}{{z}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{2}}\]
We will now rearrange these terms to give us:
\[1+{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{2}}\] …(1)
This is the simplification of the first term.
Now we will simplify the second term and check whether that equals equation (1).
Let us first simplify the second term which is ${{\left( 1-\sum{xy} \right)}^{2}}+{{\left( \sum{x-xyz} \right)}^{2}}$
We know that\[{{\left( a-b \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\].
 Using this formula, we will open and simplify the two brackets.
On opening the brackets, we get the following:
 \[\left[ {{1}^{2}}+{{\left( \sum{xy} \right)}^{2}}-2\times 1\times \left( \sum{xy} \right) \right]+\left[ {{\left( \sum{x} \right)}^{2}}+{{\left( xyz \right)}^{2}}-2\times xyz\times \left( \sum{x} \right) \right]\] …(2)
We will now evaluate\[{{\left( \sum{xy} \right)}^{2}}\],\[2\times 1\times \left( \sum{xy} \right)\] ,\[{{\left( \sum{x} \right)}^{2}}\] , and \[2\times xyz\times \left( \sum{x} \right)\] separately and substitute these values in equation (2).
First, we will evaluate \[{{\left( \sum{xy} \right)}^{2}}\]
 \[{{\left( \sum{xy} \right)}^{2}}={{\left( xy+yz+xz \right)}^{2}}\]
We know that\[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\].
Using this formula, we get the following:
\[{{\left( \sum{xy} \right)}^{2}}={{\left( xy \right)}^{2}}+{{\left( yz \right)}^{2}}+{{\left( xz \right)}^{2}}+2\times xy\times yz+2\times yz\times xz+2\times xy\times xz\]
\[{{\left( \sum{xy} \right)}^{2}}={{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{z}^{2}}+2x{{y}^{2}}z+2xy{{z}^{2}}+2{{x}^{2}}yz\] …(3)
Next, we will evaluate \[2\times 1\times \left( \sum{xy} \right)\]
\[2\times 1\times \left( \sum{xy} \right)=2\times \left( xy+yz+xz \right)\]
\[2\times 1\times \left( \sum{xy} \right)=2xy+2yz+2xz\] …(4)
Next, we will evaluate \[{{\left( \sum{x} \right)}^{2}}\]
\[{{\left( \sum{x} \right)}^{2}}={{\left( x+y+z \right)}^{2}}\]
We know that\[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\].
Using this formula, we get the following:
\[{{\left( \sum{x} \right)}^{2}}={{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz\] …(5)
Now, we will evaluate \[2\times xyz\times \left( \sum{x} \right)\]
 \[2\times xyz\times \left( \sum{x} \right)=2\times xyz\times \left( x+y+z \right)\]
\[2\times xyz\times \left( \sum{x} \right)=2{{x}^{2}}yz+2x{{y}^{2}}z+2xy{{z}^{2}}\] …(6)
Now, we will substitute equations (3), (4), (5), and (6) in equation (2)
\[\left[ {{1}^{2}}+{{\left( \sum{xy} \right)}^{2}}-2\times 1\times \left( \sum{xy} \right) \right]+\left[ {{\left( \sum{x} \right)}^{2}}+{{\left( xyz \right)}^{2}}-2\times xyz\times \left( \sum{x} \right) \right]=\]
\[\left[ 1+\left( {{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{z}^{2}}+2x{{y}^{2}}z+2xy{{z}^{2}}+2{{x}^{2}}yz \right)-\left( 2xy+2yz+2xz \right) \right]+\] \[\left[ {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz+{{x}^{2}}{{y}^{2}}{{z}^{2}}-\left( 2{{x}^{2}}yz+2x{{y}^{2}}z+2xy{{z}^{2}} \right) \right]\]
\[\begin{align}
  & \Rightarrow 1+{{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{z}^{2}}+2x{{y}^{2}}z+2xy{{z}^{2}}+2{{x}^{2}}yz-2xy-2yz-2xz+ \\
 & {{x}^{2}}+{{y}^{2}}+{{z}^{2}}+2xy+2yz+2xz+{{x}^{2}}{{y}^{2}}{{z}^{2}}-2{{x}^{2}}yz-2x{{y}^{2}}z-2xy{{z}^{2}} \\
\end{align}\]
Here, we see that many elements are cancelled out. These elements are:
\[2xy,2yz,2xz,2{{x}^{2}}yz,2x{{y}^{2}}z,2xy{{z}^{2}}\]
So we remove these elements to get the following:
\[\left[ {{1}^{2}}+{{\left( \sum{xy} \right)}^{2}}-2\times 1\times \left( \sum{xy} \right) \right]+\left[ {{\left( \sum{x} \right)}^{2}}+{{\left( xyz \right)}^{2}}-2\times xyz\times \left( \sum{x} \right) \right]=\]
\[1+{{x}^{2}}{{y}^{2}}+{{y}^{2}}{{z}^{2}}+{{x}^{2}}{{z}^{2}}+{{x}^{2}}+{{y}^{2}}+{{z}^{2}}+{{x}^{2}}{{y}^{2}}{{z}^{2}}\]
This is the same as the expression in equation (1)
So, $\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$ and ${{\left( 1-\sum{xy} \right)}^{2}}+{{\left( \sum{x-xyz} \right)}^{2}}$ can be simplified to the same form.
And hence, $\left( 1+{{x}^{2}} \right)\left( 1+{{y}^{2}} \right)\left( 1+{{z}^{2}} \right)$can be expressed as${{\left( 1-\sum{xy} \right)}^{2}}+{{\left( \sum{x-xyz} \right)}^{2}}$.
And so the answer is 1.

Note: This question involves a lot of long calculations. So you have to approach it carefully. Use the formulae \[{{\left( a-b \right)}^{2}}=\left( {{a}^{2}}+{{b}^{2}}-2ab \right)\] and \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ac\] to simplify the calculation instead of actually multiplying two or three times.
Simplify both the terms to reach a common expression instead of trying to modify one term to get the other term as it will get very difficult to group terms to get the desired expression.