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 remember the formula of difference of cubes, that says: x^3-y^3=(x-y)(x^2+xy+y^2).

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-4)(x^2+4x+16)

2018年7月3日 · We can solve this without these: Notice we can write: 27=3^3 64=4^3 :. (3/4)^x=(3^3)/(4^3) By the laws of indices: (3^3)/(4^3)=(3/4)^3 :. (3/4)^x=(3/4)^3 Since both bases are the same, both exponents are equal: :. x=3

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 ...

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 more obvious one stands out to me, because we have an #x^2# with an #x^3# on the numerator.

2015年5月23日 · (4/3)^x=27/64 (4/3)^x=(3/4)^3 upon taking the reciprocal of the second term the sign of the exponent will reverse from positive to negative. (4/3)^x=(4/3)^-3 now we can equate the exponents for the expression: x = -3 is the solution.

2018年5月9日 · x^3=64 >"using the "color(blue)"law of logarithms" •color(white)(x)log_b x=nhArrx=b^n "here "x=64,b=x" and "n=3 rArrlog_x 64=3rArrx^3=64

2015年12月9日 · I found x=64 You can find x by changing the log into an exponential taking the base, 4, from the left and "pushing up" 3 on the right: log_4x=3 so: x=4^3=64 hope it ...

2016年10月4日 · This is the difference of two cubes, ie X^3-Y^3 = (X-Y)(X^2+XY+Y^2) (Learn This) 125x^3−64y^3=(5x)^3-(4y)^3 =(5x-4y)((5x)^2+(5x)(4y)+(4y)^2) =(5x-4y)(25x^2+20xy+16y^2)

2015年4月15日 · If you write #x^6-64 = (x^2)^3 - 4^3# you can factor it into: #(x+2)(x-2)(x^4 +4x^2+16)# If you write: #x^6-64 = (x^3)^2 - 8^2#, then you factor as #(x^3+8)(x^3-8)# which can be further factored as: #(x+2)(x^2-2x+4)(x-2)(x^2+2x+4)# The second answer is factored into irreducibles (over #RR#).