
Factorise: \[{x^6} - 729\].
Answer
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Hint: Here, we are required to factorize the given expression. We will first write the numeral value present in the expression in the form of exponents. We will then simplify it using the properties of exponents and write 729 as an exponential number with a power 6. Then we will factorize the expression using suitable algebraic identities to get the required answer.
Formula Used:
We will use the following formulas:
1. \[{\left( {{a^m}} \right)^n} = {a^{m \cdot n}}\]
2. \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
3. \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]
Complete step-by-step answer:
In order to factorise \[{x^6} - 729\],
First, we will factorize729 using the method of the prime factorization.
Therefore, 729 can be written as:
\[729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3\]
Hence, this can be written as:
\[729 = {3^6}\]
Therefore, we are required to factorise \[{x^6} - {3^6}\]
Using the formula \[{\left( {{a^m}} \right)^n} = {a^{m \cdot n}}\], we can write this as:
\[{x^6} - 729 = {\left( {{x^2}} \right)^3} - {\left( {{3^2}} \right)^3}\]
\[ \Rightarrow {x^6} - 729 = {\left( {{x^2}} \right)^3} - {\left( 9 \right)^3}\]
Now, using the formula, \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\], we get,
\[ \Rightarrow {x^6} - 729 = \left( {{x^2} - 9} \right)\left[ {{{\left( {{x^2}} \right)}^2} + {{\left( 9 \right)}^2} + \left( {{x^2}} \right)\left( 9 \right)} \right]\]
Applying the exponent on terms, we get
\[ \Rightarrow {x^6} - 729 = \left( {{x^2} - {3^2}} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Using the formula, \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the first bracket, we get
\[ \Rightarrow {x^6} - 729 = \left( {x - 3} \right)\left( {x + 3} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Therefore, \[{x^6} - 729\] can be factored as \[\left( {x - 3} \right)\left( {x + 3} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Hence, this is the required answer.
Note:
When a number is written with a power then the power becomes the exponent of that particular number. It shows how many times that particular number will be multiplied by itself. We know that factorization is a method of writing an original number as the product of its various factors. Also, prime numbers are those numbers which are greater than 1 and have only two factors, i.e. factor 1 and the prime number itself. Hence, prime factorization is a method in which we write the original number as the product of various prime numbers.
Formula Used:
We will use the following formulas:
1. \[{\left( {{a^m}} \right)^n} = {a^{m \cdot n}}\]
2. \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\]
3. \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\]
Complete step-by-step answer:
In order to factorise \[{x^6} - 729\],
First, we will factorize729 using the method of the prime factorization.
Therefore, 729 can be written as:
\[729 = 3 \times 3 \times 3 \times 3 \times 3 \times 3\]
Hence, this can be written as:
\[729 = {3^6}\]
Therefore, we are required to factorise \[{x^6} - {3^6}\]
Using the formula \[{\left( {{a^m}} \right)^n} = {a^{m \cdot n}}\], we can write this as:
\[{x^6} - 729 = {\left( {{x^2}} \right)^3} - {\left( {{3^2}} \right)^3}\]
\[ \Rightarrow {x^6} - 729 = {\left( {{x^2}} \right)^3} - {\left( 9 \right)^3}\]
Now, using the formula, \[{a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + {b^2} + ab} \right)\], we get,
\[ \Rightarrow {x^6} - 729 = \left( {{x^2} - 9} \right)\left[ {{{\left( {{x^2}} \right)}^2} + {{\left( 9 \right)}^2} + \left( {{x^2}} \right)\left( 9 \right)} \right]\]
Applying the exponent on terms, we get
\[ \Rightarrow {x^6} - 729 = \left( {{x^2} - {3^2}} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Using the formula, \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the first bracket, we get
\[ \Rightarrow {x^6} - 729 = \left( {x - 3} \right)\left( {x + 3} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Therefore, \[{x^6} - 729\] can be factored as \[\left( {x - 3} \right)\left( {x + 3} \right)\left( {{x^4} + 9{x^2} + 81} \right)\]
Hence, this is the required answer.
Note:
When a number is written with a power then the power becomes the exponent of that particular number. It shows how many times that particular number will be multiplied by itself. We know that factorization is a method of writing an original number as the product of its various factors. Also, prime numbers are those numbers which are greater than 1 and have only two factors, i.e. factor 1 and the prime number itself. Hence, prime factorization is a method in which we write the original number as the product of various prime numbers.
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