
How do you factor $ 125{x^3} - 27 $ ?
Answer
549k+ views
Hint:In this question we need to find the factor for $ 125{x^3} - 27 $ . Here, we will rewrite the given equation, as the numbers are perfect cubes. Then, it will be converted into the form of difference of cubes. Here, we will apply the formula of difference of cubes. And, substitute the value of $ a $ and $ b $ , by which we will get the required factor of the given.
Complete step-by-step solution:
Now, we need to find the factor for $ 125{x^3} - 27 $ .
We can see here that both $ 125 $ and $ 27 $ are perfect cubes.
Therefore, we can rewrite $ 125{x^3} - 27 $ as,
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} $
Now, let us factor using the difference of cubes formula,
$ {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) $
Here, $ a = 5x $ and $ b = 3 $
Now, by substituting the values, we have,
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} = \left( {5x - 3} \right)\left( {{{\left( {5x} \right)}^2} + 5x
\times 3 + {{\left( 3 \right)}^2}} \right) $
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} $ $ = \left( {5x - 3} \right)\left( {25{x^2} + 15x + 9} \right) $
Hence, the factors of $ 125{x^3} - 27 $ are $ \left( {5x - 3} \right) $ and $ \left( {25{x^2} + 15x + 9} \right) $ .
Note: It is important to note here that when we are facing a cubic equation, first group the polynomial into two sections. Find what is common in both sections. Then, factor the commonalities out of the two terms, if each of the two terms contains the same factor, which is the required solution. However here we have used the difference of the cubes formula as the numbers are in perfect cube, we can’t use this method for all cubic equations. A cubic polynomial is a polynomial of the form, $ a{x^3} + b{x^2} + cx + d $ where $ a $ is non-zero. Normally, the cubic equation can be solved in other ways too, which depends on the polynomial given. However, with a little practice solving the cubic equation becomes a piece of cake. Also, being aware of the formulas helps us solve these kinds of problems.
Complete step-by-step solution:
Now, we need to find the factor for $ 125{x^3} - 27 $ .
We can see here that both $ 125 $ and $ 27 $ are perfect cubes.
Therefore, we can rewrite $ 125{x^3} - 27 $ as,
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} $
Now, let us factor using the difference of cubes formula,
$ {a^3} - {b^3} = \left( {a - b} \right)\left( {{a^2} + ab + {b^2}} \right) $
Here, $ a = 5x $ and $ b = 3 $
Now, by substituting the values, we have,
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} = \left( {5x - 3} \right)\left( {{{\left( {5x} \right)}^2} + 5x
\times 3 + {{\left( 3 \right)}^2}} \right) $
$ {\left( {5x} \right)^3} - {\left( 3 \right)^3} $ $ = \left( {5x - 3} \right)\left( {25{x^2} + 15x + 9} \right) $
Hence, the factors of $ 125{x^3} - 27 $ are $ \left( {5x - 3} \right) $ and $ \left( {25{x^2} + 15x + 9} \right) $ .
Note: It is important to note here that when we are facing a cubic equation, first group the polynomial into two sections. Find what is common in both sections. Then, factor the commonalities out of the two terms, if each of the two terms contains the same factor, which is the required solution. However here we have used the difference of the cubes formula as the numbers are in perfect cube, we can’t use this method for all cubic equations. A cubic polynomial is a polynomial of the form, $ a{x^3} + b{x^2} + cx + d $ where $ a $ is non-zero. Normally, the cubic equation can be solved in other ways too, which depends on the polynomial given. However, with a little practice solving the cubic equation becomes a piece of cake. Also, being aware of the formulas helps us solve these kinds of problems.
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