Question

How do you simplify $ \dfrac{{{x}^{3}}-1}{x-1} $ ?

Answer 1

Hint: In this question, we need to simplify the given expression $ \dfrac{{{x}^{3}}-1}{x-1} $ . For this, we will try to factorise the numerator of the expression using the property of algebra. Then we will cancel out the common factor from the numerator and the denominator. This will give us our final simplified expression. Property of algebra that we will use is, the difference of the cubes of two numbers a and b is given as, $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ .

Complete step by step answer:
$\Rightarrow$ Here we are given the expression as $ \dfrac{{{x}^{3}}-1}{x-1} $ . We need to simplify it. For this, let us first try to factorise the numerator of the expression. We have the numerator as $ {{x}^{3}}-1 $ . We know $ {{\left( 1 \right)}^{3}} $ is equal to 1 so we can write it as $ {{x}^{3}}-{{\left( 1 \right)}^{3}} $ .
$\Rightarrow$ As we can see, it is of the form of subtracting the cubes of two numbers. We know formula for subtracting the cubes of two numbers is given by $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right) $ . So here a = x and b = 1 we get,
 $ {{x}^{3}}-{{1}^{3}}=\left( x-1 \right)\left( {{x}^{2}}+x\left( 1 \right)+{{1}^{2}} \right)\Rightarrow {{x}^{3}}-{{1}^{3}}=\left( x-1 \right)\left( {{x}^{2}}+x+{{1}^{2}} \right) $ .
$\Rightarrow$ Let us rewrite $ {{\left( 1 \right)}^{3}} $ as 1 we get $ {{x}^{3}}-1=\left( x-1 \right)\left( {{x}^{2}}+x+{{1}^{2}} \right) $ .
$\Rightarrow$ Now let us put this value of the numerator of the original expression we get, $ \dfrac{\left( x-1 \right)\left( {{x}^{2}}+x+{{1}^{2}} \right)}{\left( x-1 \right)} $ .
$\Rightarrow$ As we can see (x-1) is a factor common to both the numerator and the denominator of the expression. So we can cancel it. Thus our expression reduces to $ \left( {{x}^{2}}+x+{{1}^{2}} \right) $ .
$\Rightarrow$ This expression cannot be simplified further so our required expression is $ \left( {{x}^{2}}+x+{{1}^{2}} \right) $ .

Note:
Students often get confused between the two identities that are $ {{a}^{3}}-{{b}^{3}},{{a}^{3}}+{{b}^{3}} $ . They can make mistakes of sign in these. Students should note that $ {{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\text{ and }{{a}^{3}}+{{b}^{3}}=\left( a+b \right)\left( {{a}^{2}}-ab+{{b}^{2}} \right) $ we can cancel only common factor from the numerator and the denominator.