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 …

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) = …

(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 rArr27x^3 ...

2015年12月11日 · = (3x-5) (9x^"2" +15x+25) you can remember to factor difference between 2 cubes as this a^"3" - b^"3" = (a-b) (a^2 + ab + b^2) so the answer to your question.. …

2018年5月15日 · (3x+2y)(9x^2-6xy+4y^2) 27x^3+8y^3 =(3x)^3+(2y)^3 =(3x+2y)((3x)^2-(3x)(2y)+(2y)^2) =(3x+2y)(9x^2-6xy+4y^2)

2016年4月26日 · 27x^3-125y^3=(3x-5y)(9x^2+15xy+25y^2) Notice that both of the terms are perfect cubes: 27x^3 = (3x)^3 125y^3 = (5y)^3 So we can conveniently use the difference of …

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.

2015年5月28日 · Use the identity: a^3 + b^3 = (a + b)(a^2 - ab + b^2) 27x^3 + 1^3 = (3x + 1)(9x^2 - 3x + 1) How do you find the two numbers by using the factoring method, if one number is …

2016年10月5日 · 27x^3+8=(3x)^3+2^3=(3x+2)(9x^2-6x+4) Think of 27x^3 as (3x)^3 and 8 as 2^3. This is a sum of cubes and we can apply the sum of cubes formula: a^3 + b^3 = (a + b) (a^2 − …