Question

Factorise each of the following: $ 1 - 64{a^3} - 12a + 48{a^2} $

Answer 1

Hint: In this expression we can find the factors by using the identity and compare the given expression with that identity. Then by substituting values get the simple expression which is called factors of given expression.

Complete step-by-step answer:
The given expression is $ 1 - 64{a^3} - 12a + 48{a^2} $ .
The given expression has the degree of 3 and we know that the identity of whole cube over two number which have negative operation between them is given as,
 $ {\left( {a - b} \right)^3} = {a^3} - {b^3} - 3{a^2}b + 3a{b^2} $
Here, in the given we can see that 64 is the cube of 4 and 1 is the cube of 1 so taking the given expression as:
 $ {\left( 1 \right)^3} - {\left( {4a} \right)^3} - 3 \times {\left( 1 \right)^2} \times 4a + 3 \times 1 \times {\left( {12a} \right)^2} $
If we compare the expression with the identity mentioned above then we get,
 $ a = 1,b = 4a $
Here we get the value of a and b so we can take the values of a and b to factorise or get the factor of the given expression,
 $
\Rightarrow {\left( 1 \right)^3} - {\left( {4a} \right)^3} - 3 \times {\left( 1 \right)^2} \times 4a + 3 \times 1 \times {\left( {12a} \right)^2}\\
  {\left( {1 - 4a} \right)^3}
 $
After substituting the values of the expression in the identity then we get the factorise from the expression which is the exponent of 3. Taking the above as:
 $ \Rightarrow \left( {1 - 4a} \right)\left( {1 - 4a} \right)\left( {1 - 4a} \right) $
This is the simplest expression which is the factor of the given expression which means the given expression can be completely divisible by these factors.

Note: This expression can also be simplified or we can find the factor of the given expression by taking common from the given expression. You have to take out the factor or make the expression in simple till when there should be no possibility of taking out multiple forms. These simplest values are called the factor of the given expression as they can divide the expression completely.