
How do you factor \[125{x^3} + 64\]?
Answer
555k+ views
Hint: Here, we will express each term as a cube of appropriate factors. We will do this by prime factorization. Then, we will apply a suitable algebraic identity to resolve the expression into a product of factors.
Formula used:
We will use the following formulas:
1. \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]
2. \[{a^m}{b^m} = {(ab)^m}\]
Complete step by step solution:
We are required to factor the expression \[125{x^3} + 64\]. To do this, let us begin by finding the prime factorization of each term and observe how we can express these terms as cubes of appropriate factors.
First, let us find the prime factorization of \[125\].
We know that \[125\] can be written as \[125 = 5 \times 5 \times 5\].
We see that the prime factorization of \[125\] gives us 5 multiplied to itself three times. Any number multiplied by itself three times is called a cube.
So, we can write \[125\] as a cube of 5 i.e., \[125 = {5^3}\]
Now, in the same term, we have \[{x^3}\], which is \[x\] multiplied by itself three times.
In the first term, we have \[125{x^3}\] and we have \[125 = {5^3}\], so we can write this term as a product of two cubes i.e.,
\[125{x^3} = {5^3} \times {x^3}\]
Using the law of exponent \[{a^m}{b^m} = {(ab)^m}\] to the RHS of the above expression, we get
\[ \Rightarrow 125{x^3} = {(5x)^3}\]
Now, we have to write the term \[64\] as a cube of a factor.
We see that
\[64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]
\[ \Rightarrow 64 = 4 \times 4 \times 4\].
Hence, we can write \[64\] as a cube of 4 i.e.,
\[ \Rightarrow 64 = {4^3}\]
Thus, the expression \[125{x^3} + 64\] can be written as
\[125{x^3} + 64 = {(5x)^3} + {4^3}\] ………\[(1)\]
Now, let us apply the identity \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\] to the RHS of equation \[(1)\], where \[a = 5x\] and \[b = 4\]. Hence, we have
\[125{x^3} + 64 = (5x + 4)({(5x)^2} - (5x)(4) + {4^2})\]
Multiplying the terms, we get
\[ \Rightarrow 125{x^3} + 64 = (5x + 4)(25{x^2} - 20x + 16)\]
Note:
In the identity \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], it is worth observing that the first factor has a ‘\[ + \]’ . On the other hand, the identity \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\] has a ‘\[ - \]’ in the first factor. Also, we need to keep in mind that the identities \[{a^3} + {b^3}\] and \[{(a + b)^3}\] are not the same. In \[{a^3} + {b^3}\], each term is cubed and then added whereas in \[{(a + b)^3}\], the terms are added and then cubed.
Formula used:
We will use the following formulas:
1. \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\]
2. \[{a^m}{b^m} = {(ab)^m}\]
Complete step by step solution:
We are required to factor the expression \[125{x^3} + 64\]. To do this, let us begin by finding the prime factorization of each term and observe how we can express these terms as cubes of appropriate factors.
First, let us find the prime factorization of \[125\].
We know that \[125\] can be written as \[125 = 5 \times 5 \times 5\].
We see that the prime factorization of \[125\] gives us 5 multiplied to itself three times. Any number multiplied by itself three times is called a cube.
So, we can write \[125\] as a cube of 5 i.e., \[125 = {5^3}\]
Now, in the same term, we have \[{x^3}\], which is \[x\] multiplied by itself three times.
In the first term, we have \[125{x^3}\] and we have \[125 = {5^3}\], so we can write this term as a product of two cubes i.e.,
\[125{x^3} = {5^3} \times {x^3}\]
Using the law of exponent \[{a^m}{b^m} = {(ab)^m}\] to the RHS of the above expression, we get
\[ \Rightarrow 125{x^3} = {(5x)^3}\]
Now, we have to write the term \[64\] as a cube of a factor.
We see that
\[64 = 2 \times 2 \times 2 \times 2 \times 2 \times 2\]
\[ \Rightarrow 64 = 4 \times 4 \times 4\].
Hence, we can write \[64\] as a cube of 4 i.e.,
\[ \Rightarrow 64 = {4^3}\]
Thus, the expression \[125{x^3} + 64\] can be written as
\[125{x^3} + 64 = {(5x)^3} + {4^3}\] ………\[(1)\]
Now, let us apply the identity \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\] to the RHS of equation \[(1)\], where \[a = 5x\] and \[b = 4\]. Hence, we have
\[125{x^3} + 64 = (5x + 4)({(5x)^2} - (5x)(4) + {4^2})\]
Multiplying the terms, we get
\[ \Rightarrow 125{x^3} + 64 = (5x + 4)(25{x^2} - 20x + 16)\]
Note:
In the identity \[{a^3} + {b^3} = (a + b)({a^2} - ab + {b^2})\], it is worth observing that the first factor has a ‘\[ + \]’ . On the other hand, the identity \[{a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})\] has a ‘\[ - \]’ in the first factor. Also, we need to keep in mind that the identities \[{a^3} + {b^3}\] and \[{(a + b)^3}\] are not the same. In \[{a^3} + {b^3}\], each term is cubed and then added whereas in \[{(a + b)^3}\], the terms are added and then cubed.
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