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Can someone please help me?

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

2 Answers

6 votes

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}

User Keith Grout
by
4.9k points
6 votes

Answer:


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

User Evan Priestley
by
5.4k points
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