Question

Can someone please help me?

`\frac{64x^(3)-{1} }{4x-1}`

Answer 1

`\bf{= \frac{64 {x}^(3) - 1 }{4x - 1} }\\ \\ \bf{= \frac{( \cancel{4x - 1})(16 {x}^(2) + 4x + 1) }{ \cancel{4x - 1}}} \\\\ \bf{= 16 {x}^(2) + 4x + 1}`

Answer 2

`16x^2+4x+1`

Explanation:

Given expression:

`(64x^3-1)/(4x-1)`

Step 1

Factor the numerator of the given expression.

Rewrite 64 as 4³ and 1 as 1³:

`\implies (4^3)x^3-1^3`

`\textsf{Apply the exponent rule} \quad a^b \cdot c^b=(ac)^b:`

`\implies (4x)^3-1^3`

`\textsf{Apply the Difference of Cubes Formula} \quad x^3-y^3=\left(x-y\right)\left(x^2+xy+y^2\right):`

`\implies (4x)^3-1^3=(4x-1)\left((4x)^2+4x(1)+(1)^2\right)`

`\implies (4x)^3-1^3=(4x-1)\left(16x^2+4x+1\right)`

Step 2

Replace the numerator in the given expression with the factored numerator from step 1:

`\implies (64x^3-1)/(4x-1)=((4x-1)\left(16x^2+4x+1\right))/(4x-1)`

Cancel the common factor (4x - 1):

`\implies 16x^2+4x+1`