How do you factor ${x^3} + \dfrac{1}{8}$?
Answer
623.4k+ views
Hint: We will first write the formula of ${a^3} + {b^3}$, then we will find the values of a and b according to the given and the mentioned formula and then just put in the values.
Complete step-by-step solution:
We are given that we need to factor ${x^3} + \dfrac{1}{8}$.
We know that we have formula given by the following expression:-
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
Putting a = x and $b = \dfrac{1}{2}$ in the above mentioned formula which is given by the expression ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$, we will then obtain the following expression with us:-
$ \Rightarrow {x^3} + {\left( {\dfrac{1}{2}} \right)^3} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + {{\left( {\dfrac{1}{2}} \right)}^2} - x \times \left( {\dfrac{1}{2}} \right)} \right\}$
Simplifying the calculations in the curly bracket on the right hand side of the above expression, we will then obtain the following expression with us:-
$ \Rightarrow {x^3} + {\left( {\dfrac{1}{2}} \right)^3} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Simplifying the left hand side of the above expression, we will then obtain:-
$ \Rightarrow {x^3} + \dfrac{1}{8} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Hence, the required answer is:-
$ \Rightarrow {x^3} + \dfrac{1}{8} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Note: The students must note that they must commit to memory the formula given by the following expression:-
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
We used this formula in the solution given above. But we need to make sure that it is right.
We can also verify this formula. Because since we used it, we must know its truth value because if it is just a vague formula, then we cannot use it. Therefore, it must be proved somehow as well.
Consider the right hand side of the formula which is given by the expression which is: $\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
Now, we will use the fact that x (u + v + w) = xu + xv + xz
We will therefore obtain the following expression:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = a\left( {{a^2} + {b^2} - ab} \right) + b\left( {{a^2} + {b^2} - ab} \right)$
Simplifying it further, we will then obtain the following expression with us:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + a{b^2} - {a^2}b + {a^2}b + {b^3} - a{b^2}$
Simplifying it further, we will then obtain the following expression with us:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + {b^3}$
This is equal to the left hand side.
Hence L H S = R H S.
Therefore, the formula given by the expression $\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + {b^3}$ we used is accurate.
Complete step-by-step solution:
We are given that we need to factor ${x^3} + \dfrac{1}{8}$.
We know that we have formula given by the following expression:-
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
Putting a = x and $b = \dfrac{1}{2}$ in the above mentioned formula which is given by the expression ${a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$, we will then obtain the following expression with us:-
$ \Rightarrow {x^3} + {\left( {\dfrac{1}{2}} \right)^3} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + {{\left( {\dfrac{1}{2}} \right)}^2} - x \times \left( {\dfrac{1}{2}} \right)} \right\}$
Simplifying the calculations in the curly bracket on the right hand side of the above expression, we will then obtain the following expression with us:-
$ \Rightarrow {x^3} + {\left( {\dfrac{1}{2}} \right)^3} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Simplifying the left hand side of the above expression, we will then obtain:-
$ \Rightarrow {x^3} + \dfrac{1}{8} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Hence, the required answer is:-
$ \Rightarrow {x^3} + \dfrac{1}{8} = \left( {x + \dfrac{1}{2}} \right)\left\{ {{x^2} + \dfrac{1}{4} - \dfrac{x}{2}} \right\}$
Note: The students must note that they must commit to memory the formula given by the following expression:-
$ \Rightarrow {a^3} + {b^3} = \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
We used this formula in the solution given above. But we need to make sure that it is right.
We can also verify this formula. Because since we used it, we must know its truth value because if it is just a vague formula, then we cannot use it. Therefore, it must be proved somehow as well.
Consider the right hand side of the formula which is given by the expression which is: $\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right)$
Now, we will use the fact that x (u + v + w) = xu + xv + xz
We will therefore obtain the following expression:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = a\left( {{a^2} + {b^2} - ab} \right) + b\left( {{a^2} + {b^2} - ab} \right)$
Simplifying it further, we will then obtain the following expression with us:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + a{b^2} - {a^2}b + {a^2}b + {b^3} - a{b^2}$
Simplifying it further, we will then obtain the following expression with us:-
$ \Rightarrow \left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + {b^3}$
This is equal to the left hand side.
Hence L H S = R H S.
Therefore, the formula given by the expression $\left( {a + b} \right)\left( {{a^2} + {b^2} - ab} \right) = {a^3} + {b^3}$ we used is accurate.
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