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# How do you factor completely ${x^6} - {y^6}$.

Last updated date: 17th Jun 2024
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Hint: In this question, we used three formulas for factor. First formula: the power of ‘a’ is two $\left( 2 \right)$ minus power of ‘b’ is two. The expression of the power of ‘a’ is two minus the power of ‘b’ is two is called the difference of squares.
And this formula is expressed as below.
$\left( {{a^2} - {b^2}} \right) = \left( {a - b} \right)\left( {a + b} \right)$
This is the formula for the difference of squares.

Complete step by step solution:
We know that, we find the factor of $\left( {{a^2} - {b^2}} \right)$ form type.
First, we want to calculate the factor of the above formula.
Then,
$\Rightarrow {a^2} - {b^2}$
In the above expression, we can add and subtract the $ab$. By adding and subtracting the $ab$, there is no effect in the above expression.
Then,
The above expression is written as below.
$\Rightarrow {a^2} - {b^2} + ab - ab$
This is written as below form.
${a^2} - ab + ab - {b^2}$
Then we take the common, in the first two we take $a$ as common and in the last two we take the $b$ as common.
Then the above expression is written as below.
$a\left( {a - b} \right) + b\left( {a - b} \right)$
By solving the above expression, the result would be as below.
$\left( {a - b} \right)\left( {a + b} \right)$
Then, we prove that the ${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
And also we can find the factor of ${a^3} - {b^3}$and ${a^3} + {b^3}$as given below.
The factor of ${a^3} - {b^3}$is.
${a^3} - {b^3} = \left( {a - b} \right).\left( {{a^2} + ab + {b^2}} \right)$
And the factor of ${a^3} + {b^3}$ is.
${a^3} + {b^3} = \left( {a + b} \right).\left( {{a^2} - ab + {b^2}} \right)$
Let’s come to the question, in the question the data is given below.
${x^6} - {y^6}$
We can write the above equation as below.
$\Rightarrow {\left( {{x^3}} \right)^2} - {\left( {{y^3}} \right)^2}$
We know that,
${a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)$
Then we find the factor of above as below.
$\Rightarrow {\left( {{x^3}} \right)^2} - {\left( {{y^3}} \right)^2} = \left( {{x^3} - {y^3}} \right)\left( {{x^3} + {y^3}} \right)..........\left( 1 \right)$
We know that,
${a^3} - {b^3} = \left( {a - b} \right).\left( {{a^2} + ab + {b^2}} \right) \\ {a^3} + {b^3} = \left( {a + b} \right).\left( {{a^2} - ab + {b^2}} \right) \\$
According to these formulas, the above expression is written as below.
$= \left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$

Therefore, the factor of ${x^6} - {y^6}$ are $\left( {x - y} \right)\left( {{x^2} + xy + {y^2}} \right)\left( {x + y} \right)\left( {{x^2} - xy + {y^2}} \right)$.

Note:
In this question, we find the factor of ${a^2} - {b^2}$. The formula for finding the factor of ${a^2} - {b^2}$ type is derived by adding and subtracting the $ab$term. If the number which comes in place of $a$ and $b$ is a perfect square. Then the factor can easily find out. And also we find the factor of ${a^3} - {b^3}$and ${a^3} + {b^3}$as below.
Then we used another two formulas. And formulas are given below.
$\Rightarrow {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right)$
$\Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} - ab + {b^2}} \right)$.