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Use this formula: (a^3 + b^3) = (a^2 - a*b + b^2)
but i thought it was ax2+bx+c, so what would i use as b if theres only an A and a C
The a and b are just letters. If you want to label them a and c, that's fine. So (a^3 + c^3) = (a^2 - a*c + c^2) where a^3 = 8x^3, and b^3 = 27. Then, a = 2x and b = 3.
so (8x^3+27^3)=(8^2-8(27)+27^2) and solve?
No. (8x^3 + 27) is what you have to factor. So it would be (2x + 3)(4x^2 - 6x + 9).
I believe I wrote the original equation wrong, I'm sorry! If you are given something in the form of (a^3 + c^3), then this factors out to (a + c)*(a^2 - ac + c^2)
but the c isn't raised to the 3rd power its just 27 only the 8 is raised to the 3rd power. would that make a difference?
You are right, 27 is not raised to the third power. However, you can change 27 into 3^3, correct? You could also change 8x^3 into (2x)^3, correct? That way, you can get it in the form above! :)
ok i get it! so now i put it in that form and i get (8+3)*(8^2-8(3)+3^2) but what happened to the x! im sorry for all the questions!!
That's ok! I'll walk through it slowly this time! So we originally have (8x^3 + 27). So we have the form (a^3 + b^3). When we have this form, then (a^3 + b^3) =(a+b) * (a^2 - ab + b^2). That means 8x^3 = a^3. Taking the cube root, we see that a = 2x. That also means 27 = b^3. Taking cube root, we see that b = 3. So with a = 2x, and b = 3, plug in to that formula! (a^3 + b^3) = (a + b) * (a^2 -ab + b^2) = (2x + 3) * ( (2x)^2) - (2x)(3) - (3)^2) = (2x + 3) * (4x^2 - 6x + 9). Woot woot! Mathematics!!!
ohhhhhhhh! would you still further factor the 4x^2-6x+9?
Nope! It's impossible to factor that part. After this, you are done! :)