Question

The GCD of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$ is
A. ${{x}^{3}}-1$
B. ${{x}^{3}}+1$
C. $x+1$
D. $x-1$

Answer 1

Hint: We need to find the GCD of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$. First, we need to find the common factors of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$ from their factors’ list. Then we find the greatest common factor of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$. We can also take the simultaneous factorisation of those two polynomials to find the GCD.

Complete step by step solution:
We find the factorisation of the polynomials $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$.
For $\left( {{x}^{3}}-1 \right)$, we use the factorisation form of ${{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)$.
We replace the values as $a=x;b=1$.
Therefore, $\left( {{x}^{3}}-1 \right)=\left( x-1 \right)\left( {{x}^{2}}+x+1 \right)$.
The factorisation of ${{x}^{4}}-1$ can be achieved by the use of ${{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( a-b \right)$ where we take $a={{x}^{2}};b=1$.
Therefore, the factorisation is ${{x}^{4}}-1=\left( {{x}^{2}}-1 \right)\left( {{x}^{2}}+1 \right)$.
The factorisation of ${{x}^{2}}-1$ will be ${{x}^{2}}-1=\left( x+1 \right)\left( x-1 \right)$ taking $a=x;b=1$.
Therefore, ${{x}^{4}}-1=\left( x+1 \right)\left( x-1 \right)\left( {{x}^{2}}+1 \right)$.
The common factor of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$ is $x-1$.
The greatest common factor of $\left( {{x}^{3}}-1 \right)$ and ${{x}^{4}}-1$ is $x-1$.
The correct option is D.

Note: We need to remember that the GCD has to be only one polynomial. It is the greatest possible divisor of all the given polynomials. The factors can be in the multiplication or division form but they can’t be in the form of addition or subtraction. This applies for numbers also where if for numbers $x$ and $y$, the GCD is $a$ then the GCD of the numbers $\dfrac{x}{a}$ and $\dfrac{y}{a}$ will be 1.