Question

The GCD of \[{{x}^{2}}-2xy+{{y}^{2}}\] and \[{{x}^{4}}-{{y}^{4}}\] is:
(a) 1
(b) x + y
(c) x – y
\[\left( \text{d} \right)\text{ }{{x}^{2}}-{{y}^{2}}\]

Answer 1

Hint: To solve the given question, we will first find out what GCD is and then we will check each option one by one. During checking, we will equate the term to zero in each option and we will find in terms of y in all the options. Now, we will put this value of y in the terms given in the question. The value at which both the term will be zero will be the divisors. To find GCD, we will select the highest among them. That will be the answer to the question.

Complete step by step solution:
Before solving the question, we must know what GCD is. GCD is the short form of Greatest Common Divisor. The Greatest Common Divisor of two or more integers is the largest positive number that divides each of the integers. In our question, we have to find the GCD of \[{{x}^{2}}-2xy+{{y}^{2}}\] and \[{{x}^{4}}-{{y}^{4}}.\] If a number is GCD of both the terms \[{{x}^{2}}-2xy+{{y}^{2}}\] and \[{{x}^{4}}-{{y}^{4}}\] should be zero when that number is put in place of x and y. So, we will check each option one by one.
Option (a): 1
In this case, x = 0 and y = 0. So, we will put the respective values in the above terms. Let the first term be f(x, y) and second term be g(x, y). Thus, we have,
\[f\left( x,y \right)={{x}^{2}}-2xy+{{y}^{2}}\]
\[g\left( x,y \right)={{x}^{4}}-{{y}^{4}}\]
\[\Rightarrow f\left( 0,0 \right)={{0}^{2}}-2\left( 0 \right)\left( 0 \right)+{{0}^{2}}=0\]
\[\Rightarrow g\left( 0,0 \right)={{0}^{4}}-{{0}^{4}}=0\]
Therefore, 1 is the common divisor of both f(x, y) and g(x, y).
Option (b): x + y
Now, we will equate it to zero. Thus, we have,
\[x+y=0\]
\[\Rightarrow x=-y\]
So, we will get,
\[f\left( -y,y \right)={{\left( -y \right)}^{2}}-2\left( -y \right)\left( y \right)+{{y}^{2}}\]
\[\Rightarrow f\left( -y,y \right)={{y}^{2}}+2{{y}^{2}}+{{y}^{2}}\]
\[\Rightarrow f\left( -y,y \right)=4{{y}^{2}}\]
\[g\left( -y,y \right)={{\left( -y \right)}^{4}}-{{\left( y \right)}^{4}}\]
\[\Rightarrow g\left( -y,y \right)={{y}^{4}}-{{y}^{4}}=0\]
Both are not zero, so x + y is not the common divisor.
Option (c): x – y
Now, we will equate it to zero. Thus, we have,
\[x-y=0\]
\[\Rightarrow x=y\]
So, we will get,
\[f\left( y,y \right)={{y}^{2}}-2\left( y \right)\left( y \right)+{{y}^{2}}\]
\[\Rightarrow f\left( y,y \right)={{y}^{2}}-2{{y}^{2}}+{{y}^{2}}\]
\[\Rightarrow f\left( y,y \right)=0\]
\[g\left( y,y \right)={{\left( y \right)}^{4}}-{{\left( y \right)}^{4}}\]
\[\Rightarrow g\left( y,y \right)=0\]
Both are zero, so x – y is the common divisor.
\[\text{Option }\left( \text{d} \right):{{x}^{2}}-{{y}^{2}}\]
Now we will equate it to zero. Thus, we have, \[{{x}^{2}}={{y}^{2}}.\] So, we get,
\[f\left( x,y \right)={{x}^{2}}-2xy+{{y}^{2}}\]
\[\Rightarrow f\left( x,y \right)={{y}^{2}}-2xy+{{y}^{2}}\]
\[\Rightarrow f\left( x,y \right)=2{{y}^{2}}-2xy\ne 0\]
\[g\left( x,y \right)={{x}^{4}}-{{y}^{4}}\]
\[\Rightarrow g\left( x,y \right)={{\left( {{x}^{2}} \right)}^{2}}-{{\left( {{y}^{2}} \right)}^{2}}\]
\[\Rightarrow g\left( x,y \right)={{\left( {{y}^{2}} \right)}^{2}}-{{\left( {{y}^{2}} \right)}^{2}}=0\]
Both are not zero, so, \[{{x}^{2}}-{{y}^{2}}\] is not the common divisor. Option (a) and (c) are both satisfying the conditions but \[x-y\ge 1,\] so x – y is the GCD.
Hence, option (c) is the right answer.

Note: The alternate method to solve the question is shown. We will factorize both the terms.
\[f\left( x,y \right)={{x}^{2}}-2xy+{{y}^{2}}\]
\[\Rightarrow f\left( x,y \right)={{\left( x \right)}^{2}}-2\left( x \right)\left( y \right)+{{\left( y \right)}^{2}}\]
\[\Rightarrow f\left( x,y \right)={{\left( x-y \right)}^{2}}\]
\[\Rightarrow f\left( x,y \right)=\left( x-y \right)\left( x-y \right)\]
Similarly, we have,
\[g\left( x,y \right)={{x}^{4}}-{{y}^{4}}\]
\[\Rightarrow g\left( x,y \right)={{\left( {{x}^{2}} \right)}^{2}}-{{\left( {{y}^{2}} \right)}^{2}}\]
\[\Rightarrow g\left( x,y \right)=\left( {{x}^{2}}+{{y}^{2}} \right)\left( {{x}^{2}}-{{y}^{2}} \right)\]
\[\Rightarrow g\left( x,y \right)=\left( {{x}^{2}}+{{y}^{2}} \right)\left( x-y \right)\left( x+y \right)\]
As we can see that x – y is common to both, so x – y is the GCD.