
How do you factor $27 - {x^3}$?
Answer
533.7k+ views
Hint:
Here we can proceed by writing $27$ as the cube of any number and then we will get the number of the form ${a^3} - {b^3}$
Now we can simplify it by using the formula:
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
Complete step by step solution:
Here we are given to factorise the term $27 - {x^3}$ which means that we need to simplify it and write it in the form of the factors of it.
We know that a cube of any number is a product when it is multiplied by itself three times. For example: the cube of $1 = (1)(1)(1) = 1$
Cube of $2 = (2)(2)(2) = 8$
So we need to find the number whose cube is $27$ as we can simplify this equation in the general form ${a^3} - {b^3}$
So we know that cube of $3 = (3)(3)(3) = 27$
So we can write that $27 = {3^3}$
Hence we can write the given number whose factor is to be found as:
$27 - {x^3}$$ = {3^3} - {x^3}$
Now we know that we have the formula in the cube that when we have two cubes and we do their subtraction, we get:
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So we can similarly compare ${3^3} - {x^3}$ with ${a^3} - {b^3}$ and use the formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So we can compare and we get that as per the given equation:
$
a = 3 \\
b = x \\
$
So we get that:
${3^3} - {x^3} = (3 - x)({3^2} + 3x + {x^2})$
${3^3} - {x^3} = (3 - x)({x^2} + 3x + 9)$
Hence we have found this simplified form of the factors of the given problem. We can simplify this again if we are able to get the proper factors of the quadratic equation also. Here in the quadratic equation obtained that is $({x^2} + 3x + 9)$ we cannot simplify it further as discriminant is negative.
Note:
Here in the simplified form obtained which is $(3 - x)({x^2} + 3x + 9)$ we can also simplify the equation that is quadratic which means of the degree $2$ but its discriminant must be equal to zero.
Here it is negative as $D = {b^2} - 4ac{\text{ in }}a{x^2} + bx + c$ so here we have $D = {3^2} - 4(9) = 9 - 36 = {\text{negative}}$
So we don’t need to simplify it further.
Here we can proceed by writing $27$ as the cube of any number and then we will get the number of the form ${a^3} - {b^3}$
Now we can simplify it by using the formula:
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
Complete step by step solution:
Here we are given to factorise the term $27 - {x^3}$ which means that we need to simplify it and write it in the form of the factors of it.
We know that a cube of any number is a product when it is multiplied by itself three times. For example: the cube of $1 = (1)(1)(1) = 1$
Cube of $2 = (2)(2)(2) = 8$
So we need to find the number whose cube is $27$ as we can simplify this equation in the general form ${a^3} - {b^3}$
So we know that cube of $3 = (3)(3)(3) = 27$
So we can write that $27 = {3^3}$
Hence we can write the given number whose factor is to be found as:
$27 - {x^3}$$ = {3^3} - {x^3}$
Now we know that we have the formula in the cube that when we have two cubes and we do their subtraction, we get:
${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So we can similarly compare ${3^3} - {x^3}$ with ${a^3} - {b^3}$ and use the formula ${a^3} - {b^3} = (a - b)({a^2} + ab + {b^2})$
So we can compare and we get that as per the given equation:
$
a = 3 \\
b = x \\
$
So we get that:
${3^3} - {x^3} = (3 - x)({3^2} + 3x + {x^2})$
${3^3} - {x^3} = (3 - x)({x^2} + 3x + 9)$
Hence we have found this simplified form of the factors of the given problem. We can simplify this again if we are able to get the proper factors of the quadratic equation also. Here in the quadratic equation obtained that is $({x^2} + 3x + 9)$ we cannot simplify it further as discriminant is negative.
Note:
Here in the simplified form obtained which is $(3 - x)({x^2} + 3x + 9)$ we can also simplify the equation that is quadratic which means of the degree $2$ but its discriminant must be equal to zero.
Here it is negative as $D = {b^2} - 4ac{\text{ in }}a{x^2} + bx + c$ so here we have $D = {3^2} - 4(9) = 9 - 36 = {\text{negative}}$
So we don’t need to simplify it further.
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