Question

Factor 27x^3-64

Answer 1

The answer is \[2 \times (13x^3-32)\]

Answer 2

um

Answer 3

hint: \(27x^3-64\qquad
\begin{cases}
27\implies 3\cdot 3\cdot 3\implies 3^3\\
67\implies 4\cdot 4\cdot 4\implies 4^3
\\ \quad \\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\\
(a+b)(a^2-ab+b^2)= a^3+b^3
\end{cases}\impliedby hint\)

Answer 4

well.. \(\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
(a+b)(a^2-ab+b^2)= a^3+b^3 \\ \quad \\

a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad
(a-b)(a^2+ab+b^2)= a^3-b^3\)

Answer 5

2(13x^3-32) doesn't work because 2*13 is not 27
use jdoe's approach

Answer 6

Step 1 :
Raise x to the 3rd power

Step 2 :
Simplify (2•13x3) - 64

Step 3 :Factoring: 13x3 - 32
Theory : A difference of two perfect cubes, a3 - b3 can be factored into
(a-b) • (a2 +ab +b2)
Proof : (a-b)•(a2+ab+b2) =
a3+a2b+ab2-ba2-b2a-b3 =
a3+(a2b-ba2)+(ab2-b2a)-b3 =
a3+0+0+b3 =
a3+b3

Final result :
2 • (13x3 - 32)

Answer 7

Do you get it @Jorgeydiego?

Answer 8

well, he's sorta no here

Answer 9

I can tell lol XD

Answer 10

hmm darn I even have a typo above
just to clarify \(27x^3-64\qquad
\begin{cases}
27\implies 3\cdot 3\cdot 3\implies 3^3\\
64\implies 4\cdot 4\cdot 4\implies 4^3
\\ \quad \\
\textit{difference of cubes}
\\ \quad \\
a^3+b^3 = (a+b)(a^2-ab+b^2)\qquad
\\ \quad \\

a^3-b^3 = (a-b)(a^2+ab+b^2)\qquad

\end{cases}\impliedby hint\)

Answer 11

They had different multiple choice

Answer 12

well, forget the choices, and factor it, and see what you get :)

Answer 13

\(\bf 27x^3-64\implies 3^3x^3-4^3\implies (3x)^3-(4)^3\implies ?\)

Answer 14

A.(3x-4)(9x^2-12x+16)
B.(3x-4)(9x^2+12x+16)
C.(3x+4)(9x^2-12x+16)
D.(3x+4)(9x^2+12x+16)

Answer 15

ok.... and you"you" got as factors?

Answer 16

Yeah I think so

Answer 17

ok... what did "you" get then? :)

Answer 18

\[B. (3x-4)(9x^2+12x+16)\]

Answer 19

Thank you

Answer 20

NP :D