2018年3月25日 · (3x+2)(9x^2-6x+4) Before we start, notice this: 27x^3 and 8 are both perfect cubes of 3x and 2 respectively. 27x^3+8 Now, apply sum of cubes rule: a^3+b^3=(a+b)(a^2-ab+b^2), (3x+2)(9x^2-6x+4)

2015年4月12日 · When factorising the sum of cubes, we use the formula a^3+b^3= (a+b) (a^2-ab+b^2). In this case of 27x3+64, 27x^3 = a^3 64=b^3 Find a: 27x^3 = a^3 root3 (27x^3) = root3 (a^3) 3x = a Find b: 64=b^3 root3 64 = root3 (b^3) 4 = b Substitute a=3x and b=4 into (a+b) (a^2-ab+b^2) (3x+4) ( (3x)^2 - (3x xx4) + 4^2) = (3x+4) (9x^2 - 12x + 16) (3x+4) (9x^2 - 12x + 16) is the factorised form of 27x3+64

(3x-2) (9x^2+6x+4) >27x^3-8" is a "color (blue)"difference of cubes" •color (white) (x)a^3-b^3= (a-b) (a^2+ab+b^2) "here "a=3xto (3x)^3=27x^3 "and "b=2to (2)^3=8 ...

2015年12月29日 · 27-x^3=3^3-x^3= (3-x) (9+3x+x^2) 27-x^3 is an example of the difference of cubes, where a^3-b^3= (a-b) (a^2+ab+b^2). Rewrite 27-x^3 as 3^3-x^3, where a=3 and b=x. (3-x) (3^2+ (3·x)+x^2) = (3-x) (9+3x+x^2)

2016年9月18日 · It is a real advantage to know all the powers up to 1000 by heart! In such a case, we could go the conventional method of factoring and solving fo each factor equal to 0. However, if you recognise that both 27 and 343 are cubes, as there is only one variable, we can also write as follows: 27x3 +343 = 0 27x3 = −343 x3 = − 343 27 x …

2015年12月11日 · you can remember to factor difference between 2 cubes as this a3 −b3 = (a −b)(a2 + ab + b2) so the answer to your question.. 27x3 −125 can be written as = 33x3 −53 = (3x −5)(9x2 +15x + 25) FYI: an addition between 2 cubes can be written as this a3 +b3 = (a +b)(a2 − ab + b2) it's really easy when you remember those two equations by their patterns =)

2016年4月26日 · Notice that both of the terms are perfect cubes: 27x3 = (3x)3 125y3 = (5y)3 So we can conveniently use the difference of cubes identity: A3 − B3 = (A− B)(A2 + AB + B2) with A = 3x and B = 5y as follows: 27x3 −125y3 = (3x)3 −(5y)3 = (3x −5y)((3x)2 +(3x)(5y) +(5y)2) = (3x −5y)(9x2 + 15xy + 25y2) The remaining quadratic factor has no linear factors with Real coefficients. If we allow ...

2018年5月15日 · this is a sum of cubes which factors in general as ∙ xa3 +b3 = (a +b)(a2 − ab + b2) 27x3 = (3x)3 ⇒ a = 3x 8y3 = (2y)3 ⇒ b = 2y ⇒ 27x3 + 8y3 = (3x + 2y)((3x)2 −(3x × 2y) +(2y)2) × × × × × = (3x +2y)(9x2 −6xy + 4y2)

2016年1月18日 · (27x^3 - 1) = (3x - 1) (9x^2 + 3x + 1) Cube root of 27 is 3 and the cube root of -1 is also -1. 27x^3 - 1 = (3x)^3 - (1)^3 To factor y = 27x^3 - 1, remember that (a^3 ...

2016年10月5日 · Think of 27x3 as (3x)3 and 8 as 23. This is a sum of cubes and we can apply the sum of cubes formula: a3 +b3 = (a +b)(a2 − ab + b2). Substituting a with 3x and b with 3 into the formula yields: x3 +27 = (3x + 2)(9x2 −6x + 4)