Question

Can someone please help me?
$\frac{64x^{3}-{1} }{4x-1}$

Answer 1

Answer:

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

Answer 2

Answer:

$16x^2+4x+1$

Step-by-step explanation:

Given expression:

$\dfrac{64x^3-1}{4x-1}$

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:

$\implies \dfrac{64x^3-1}{4x-1}=\dfrac{(4x-1)\left(16x^2+4x+1\right)}{4x-1}$

Cancel the common factor (4x - 1):

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