Difference of two...cubes? squares?

New member
40x^6 - 135

This was a problem on my last in-class test. I have a question regarding some of the last steps in how I worked it out. This isn't just considered prime, right? Even though none of the numbers are perfect squares. x^6 is also (x^3)^2, I know, but nothing else worked out.

So I took the GCD and got 5(8x^6 - 27). (8x^6 - 27) ends up being the difference of two cubes, right? So now I have 5 * (2x^3)^3 - (3)^3. I changed that according to rules and wrote the answer as 5(2x^3 - 3)(4x^9+6x^3+9).

I really think I got something mixed up, though, regarding the powers of the x's. What did I do wrong? Something simple, I know, in the multiplication. (As usual, I mess up because of something simple.) I appreciate your help very much!

stapel

Super Moderator
Staff member
40x^6 - 135
I will guess that the instructions were something along the lines of "factor fully".

You're nearly theres, but I think you mixed up an exponent rule: x<sup>m+m</sup> = x<sup>2m</sup> = (x<sup>2</sup>)<sup>m</sup>, not (x<sup>m</sup>)<sup>m</sup>.

. . . . .40x<sup>6</sup> - 135

. . . . .5 (8x<sup>6</sup> - 27)

. . . . .5 [(2x<sup>2</sup>)<sup>3</sup> - 3<sup>3</sup>]

. . . . .5 (2x<sup>2</sup> - 3)[(2x<sup>2</sup>)<sup>2</sup> + (2x<sup>2</sup>)(3) + 3<sup>2</sup>]

. . . . .5 (2x<sup>2</sup> - 3)(4x<sup>4</sup> + 6x<sup>2</sup> + 9)

Make sense?

Eliz.

soroban

Elite Member
Re:

Stapel is absolutely correct . . .

$$\displaystyle x^9$$ is not equal to $$\displaystyle (x^3)^3$$ . . . but rather: $$\displaystyle (x^2)^3$$

$$\displaystyle 40x^6 - 135 \;=\;5\bigg[8x^6 - 27\bigg]$$

. . . . . . . . $$\displaystyle =\;5\bigg[(2x^2)^3 - (3)^3\bigg]$$

. . . . . . . . $$\displaystyle = \;5\bigg[2x - 3)\bigg]\cdot\bigg[(2x)^2 + (2x)(3) + (3)^2\bigg]$$

. . . . . . . . $$\displaystyle = \;5(2x-3)(4x^2 + 6x + 9)$$

BTW, Eliz . . . we need brackets for exponents: x[sup:355n6b6t]3[/sup:355n6b6t]

. . But I see that: x2 doesn't work.

Subhotosh Khan

Super Moderator
Staff member
Re: Re:

soroban said:

Stapel is absolutely correct . . .

$$\displaystyle x^9$$ is not equal to $$\displaystyle (x^3)^3$$ . . . but rather: $$\displaystyle (x^2)^3$$ <--- Ahem - is that what you wanted to say

$$\displaystyle 40x^6 - 135 \;=\;5\bigg[8x^6 - 27\bigg]$$

. . . . . . . . $$\displaystyle =\;5\bigg[(2x^2)^3 - (3)^3\bigg]$$

. . . . . . . . $$\displaystyle = \;5\bigg[2x - 3)\bigg]\cdot\bigg[(2x)^2 + (2x)(3) + (3)^2\bigg]$$

. . . . . . . . $$\displaystyle = \;5(2x-3)(4x^2 + 6x + 9)$$

BTW, Eliz . . . we need brackets for exponents: x[sup:1a2niiu3]3[/sup:1a2niiu3]

. . But I see that: x2 doesn't work.

New member
Re: Re:

Subhotosh Khan said:
Stapel is absolutely correct . . .

$$\displaystyle x^9$$ is not equal to $$\displaystyle (x^3)^3$$ . . . but rather: $$\displaystyle (x^2)^3$$ <--- Ahem - is that what you wanted to say

$$\displaystyle 40x^6 - 135 \;=\;5\bigg[8x^6 - 27\bigg]$$

. . . . . . . . $$\displaystyle =\;5\bigg[(2x^2)^3 - (3)^3\bigg]$$

. . . . . . . . $$\displaystyle = \;5\bigg[2x - 3)\bigg]\cdot\bigg[(2x)^2 + (2x)(3) + (3)^2\bigg]$$

. . . . . . . . $$\displaystyle = \;5(2x-3)(4x^2 + 6x + 9)$$

Thanks to everyone who answered. I always appreciate it. I have a question though... in $$\displaystyle \;5[(2x^2)^3 - (3)^3]$$, how come we drop the squared x and it becomes (2x-3) and then the rest? I see I'm going to get this one wrong on my test... I kept the squared x when I did the last factoring steps.

Mrspi

Senior Member
Re: Re:

Subhotosh Khan said:
Stapel is absolutely correct . . .

$$\displaystyle x^9$$ is not equal to $$\displaystyle (x^3)^3$$ . . . but rather: $$\displaystyle (x^2)^3$$ <--- Ahem - is that what you wanted to say

$$\displaystyle 40x^6 - 135 \;=\;5\bigg[8x^6 - 27\bigg]$$

. . . . . . . . $$\displaystyle =\;5\bigg[(2x^2)^3 - (3)^3\bigg]$$

. . . . . . . . $$\displaystyle = \;5\bigg[2x - 3)\bigg]\cdot\bigg[(2x)^2 + (2x)(3) + (3)^2\bigg]$$

. . . . . . . . $$\displaystyle = \;5(2x-3)(4x^2 + 6x + 9)$$

Thanks to everyone who answered. I always appreciate it. I have a question though... in $$\displaystyle \;5[(2x^2)^3 - (3)^3]$$, how come we drop the squared x and it becomes (2x-3) and then the rest? I see I'm going to get this one wrong on my test... I kept the squared x when I did the last factoring steps.
You are absolutely correct....it SHOULD be

5[2x^2 - 3] [ (2x^2)^2 + 2x^2*3 + 3^2]

or,

5[2x^2 - 3] [4x^4 + 6x^2 + 9]

Probably a typo in the earlier post.