2015年2月2日 · The answer is: x^3+64=(x+4)(x^2-4x+16). It's easy if we remember the formula of sum of cubes, that says: x^3+y^3=(x+y)(x^2-xy+y^2). In this case 64=4^3. It's also to …

2016年5月11日 · x^3+64=(x+4)(x^2-4x+16) Both x^3 and 64=4^3 are perfect squares. So we can use the sum of cubes identity ...

2018年3月30日 · (x-4)(x^2+4x+16) >x^3-64larrcolor(blue)"is a difference of cubes" •color(white)(x)a^3-b^3=(a-b)(a^2+ab+b^2) "here "a=x" and "b=4to(4)^3=64 rArrx^3-64=(x …

2015年9月25日 · 2sqrt(x) (sqrt(64x^3)) ^(1/3) = (2^6*x^3)^(1/6) = 2*x^(3/6) = 2sqrt(x) 5489 views around the world . You can reuse this answer Creative Commons License

2016年1月21日 · x=-4,2+-2sqrt3i Notice that this is a sum of cubes, which is factorable as follows: a^3+b^3=(a+b)(a^2-ab+b^2) Thus, x^3+64 is factorable into x^3+4^3=(x+4)(x^2-4x+16)=0 …

64-x^3 = (4-x)(16+4x+x^2) The difference of cubes identity can be written: a^3-b^3 = (a-b)(a^2+ab+b^2) Note both 64=4^3 and x^3 are perfect cubes.

2015年5月23日 · Using the identity #a^3+b^3 = (a+b)(a^2-ab+b^2)# we find: #64x^3+27 = (4x)^3 + 3^3# #= (4x+3)((4x)^2-(4x)*3+3^2)#

2017年6月23日 · #int (x^3)/(sqrt{64+x^2}) "d"x= 1/3sqrt{64+x^2}(x^2-128) + C#. Explanation: I know you specified trigonometric substitution but I don't see why you'd use on in this case as a …

2018年5月3日 · (4x+y)(16x^2-4xy+y^2) This is an example of factoring perfect cubes. 64x^3+y^3 64 is a perfect cube 4*4*4 x^3 is a perfect cube x*x*x y^3 is a perfect cube y*y*y The factor …

2015年4月22日 · What are the factors of #x^3y^6 – 64#? How do you know if #x^2 + 10x + 25# is a perfect square? How do you write #16x^2 – 48x + 36# as a perfect square trinomial?