Question

How do you factor \[{{x}^{3}}-16\]?

Answer 1

Hint: This question is from the topic of algebra. In this question, we will first write the term 16 in cubic form so that we can make the whole term in the form of \[{{a}^{3}}-{{b}^{3}}\]. After that, we will see and use the formula of \[{{a}^{3}}-{{b}^{3}}\] in the equation which is given in the question. After that, we will solve the further process and find the factor of the term \[{{x}^{3}}-16\].

Complete step by step solution:Let us solve this question.
In this question, we have asked to find the factor of the term or we can say the equation which is given in the question that is \[{{x}^{3}}-16\].
So, let us first understand what will be the cube root of 16.
The cube root can also be written as power of 1 divided by 3.
So, we can write
\[\sqrt[3]{16}={{\left( 16 \right)}^{\dfrac{1}{3}}}\]
As we know that 16 is equal to 2 to the power of 4, so we can write
\[\Rightarrow \sqrt[3]{16}={{\left( {{2}^{4}} \right)}^{\dfrac{1}{3}}}\]
Using the formula \[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\], we can write the above equation as
\[\Rightarrow \sqrt[3]{16}={{\left( 2 \right)}^{\dfrac{4}{3}}}\]
The above equation can also be written as
\[\Rightarrow \sqrt[3]{16}={{\left( 2 \right)}^{1+\dfrac{1}{3}}}\]
Using the formula \[{{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}\], we can write the above equation as
\[\Rightarrow \sqrt[3]{16}={{\left( 2 \right)}^{1}}{{\left( 2 \right)}^{\dfrac{1}{3}}}\]
The above equation can also be written as
\[\Rightarrow \sqrt[3]{16}={{\left( 2 \right)}^{1}}\sqrt[3]{2}\]
The above equation van also be written as
\[\Rightarrow \sqrt[3]{16}=2\sqrt[3]{2}\]
As we know that \[\sqrt[n]{a}\] can also be written as \[{{a}^{\dfrac{1}{n}}}\], so we can write the above equation as
\[\Rightarrow {{16}^{\dfrac{1}{3}}}=2\sqrt[3]{2}\]
As we know that if \[{{a}^{m}}=b\], then \[a={{b}^{\dfrac{1}{m}}}\]. So, we can write the above equation as
\[\Rightarrow 16={{\left( 2\sqrt[3]{2} \right)}^{3}}\]
So, now we can the term \[{{x}^{3}}-16\] as
\[{{x}^{3}}-{{\left( 2\sqrt[3]{2} \right)}^{3}}\]
Now, using the formula \[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\], we can write the above term as
\[\left( x-2\sqrt[3]{2} \right)\left( {{x}^{2}}+2\sqrt[3]{2}x+{{\left( 2\sqrt[3]{2} \right)}^{2}} \right)\]
The above term can also be written as
\[\Rightarrow \left( x-2\sqrt[3]{2} \right)\left( {{x}^{2}}+2\sqrt[3]{2}x+4{{\left( \sqrt[3]{2} \right)}^{2}} \right)\]
The above term can also be written as
\[\Rightarrow \left( x-2\sqrt[3]{2} \right)\left( {{x}^{2}}+2\sqrt[3]{2}x+4\sqrt[3]{4} \right)\]
So, we have found the factor of the term \[{{x}^{3}}-16\]. The factor of the term \[{{x}^{3}}-16\] is \[\left( x-2\sqrt[3]{2} \right)\left( {{x}^{2}}+2\sqrt[3]{2}x+4\sqrt[3]{4} \right)\].

Note: We should have a better knowledge in the topic of algebra. we remember the following formulas to solve this type of question easily:
\[{{a}^{3}}-{{b}^{3}}=\left( a-b \right)\left( {{a}^{2}}+ab+{{b}^{2}} \right)\]
\[{{a}^{m+n}}={{a}^{m}}\times {{a}^{n}}\]
\[\sqrt[n]{a}={{a}^{\dfrac{1}{n}}}\]
\[{{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}\]
If \[{{a}^{m}}=b\], then \[a={{b}^{\dfrac{1}{m}}}\]
We should know how to find a cube root of a specific number to solve this type of question easily.