Question

How would the expression x^3-3square root of 3 be rewritten using difference of cubes?
A. (x+square root of 3)(x^2+square root of 3 +3)
B. (x-square root of 3)(x^2+3x+3)
C. (x-square root of 3)(x^2-3x-3)
D. (x-square root of 3)(x^2+square root of 3x+3)

Answer 1

@jim_thompson5910

Answer 2

Is the original expression this?
\[\Large x^3 - 3\sqrt{3}\]
or is it this?
\[\Large x^3 - \sqrt[3]{3}\]

Answer 3

the first one

Answer 4

ok let me think

Answer 5

notice how if
\[\Large y = \sqrt{3}\]
then cubing both sides gives
\[\Large y^3 = (\sqrt{3})^3\]
\[\Large y^3 = \sqrt{3}*\sqrt{3}*\sqrt{3}\]
\[\Large y^3 = \sqrt{3*3*3}\]
\[\Large y^3 = \sqrt{9*3}\]
\[\Large y^3 = \sqrt{9}*\sqrt{3}\]
\[\Large y^3 = 3\sqrt{3}\]

Answer 6

So
\[\Large x^3 - 3\sqrt{3}\]
is in the form
\[\Large x^3 - y^3\]
where
\[\Large y = \sqrt{3}\]

Answer 7

From here, use this formula
\[\Large x^3-y^3 = (x-y)(x^2+xy+y^2)\]

Answer 8

so i plug in 3 where the ys are?

Answer 9

square root of 3

Answer 10

so i just put 1.7 where the ys are right?

Answer 11

leave it in radical form

Answer 12

your answer choices have sqrt(3) in them (not 1.7)

Answer 13

ohh right x) lol ok

Answer 14

thanks

Answer 15

no problem