How do you factor x^6 - 1? - Socratic
2015年7月2日 · x^6-1 = (x^3-1)(x^3+1) =(x-1)(x^2+x+1)(x+1)(x^2-x+1) Use difference of squares, difference of cubes and sum of cubes: [1]: a^2-b^2 = (a-b)(a+b) [2]: a^3-b^3 = (a-b)(a ...
What are all the sixth roots of #1# - Socratic
2017年4月30日 · The roots of x^6 = 1 are: +-1 1/2+-sqrt(3)/2i -1/2+-sqrt(3)/2i The sixth roots of 1 are all the solutions of: x^6-1 = 0 We can solve this by factoring with the help of a few identities: Difference of squares: a^2-b^2 = (a-b)(a+b) Difference of cubes: a^3-b^3=(a-b)(a^2+ab+b^2) Sum of cubes: a^3+b^3=(a+b)(a^2-ab+b^2) Hence we find: 0 = x^6-1 ...
How do you factor 64x^6 -1? - Socratic
2016年1月19日 · Use some standard identities to find: 64x^6-1=(2x-1)(4x^2+2x+1)(2x+1)(4x^2-2x+1) The difference of squares identity can be written: a^2-b^2=(a-b)(a+b) The difference of cubes identity can be written: a^3-b^3 = (a-b)(a^2+ab+b^2) The sum of cubes identity can be written: a^3+b^3=(a+b)(a^2-ab+b^2) We find: 64x^6-1 =(8x^3)^2-1^2 =(8x^3-1)(8x^3+1 ...
How do you factor completely x^6-1? - Socratic
2016年5月13日 · The answer in the given form can be obtained directly by observing that it is a cubic in #x^2#. #x^6-1=(x^2)^3-1^3# #=(x^2-1)((x^2)^2+x^2+!)# #=(x-1)(x+1)(x^4+x^2+1)# #=(x-1)(x+1)(x^2-x+1)(x^2-x+1)# #=(x-1)(x+1)(x-1/2-i sqrt3/2)(x-1/2+i sqrt3/2)# #(x+1/2-i sqrt3/2)(x+1/2+ isqrt3/2)# The expanded form can be directly obtained from
Int 1/1+x^6 - Socratic
2018年4月3日 · #int 1/(1+x^6)dx# or. #int 1/(1+x)^6dx# Because in the 2nd case it's easier and is actually solveable whereas the 1st one is harder. So the solution to #int 1/(1+x)^6dx# is: we will take u as our new variable such as #u=(1+x)# and #du=dx# which then gives us #int 1/u^6du# and the answer is #-1/(5u^5) + # constant. so the answer is #-1/(5(1+x)^5 ...
How do you factor x^6-1? - Socratic
2016年1月2日 · Use some standard identities to find: x^6-1=(x-1)(x^2+x+1)(x+1)(x^2-x+1) Use the difference of squares identity: a^2-b^2 = (a-b)(a+b) the difference of cubes identity ...
Evaluate #int(x^6+1)/(x^2+1)dx# - Socratic
2018年3月17日 · # int (x^6+1)/(x^2+1) dx = t + tan^5t/5 - tan^3t/3 + tant - t +C# # int (x^6+1)/(x^2+1) dx = tan^5t/5 - tan^3t/3 + tant + C#. and undoing the substitution: #int (x^6+1)/(x^2+1) dx = x^5/5 -x^3/3 +x+C#. Note: from the result I realized that: #(x^6+1)/(x^2+1) = x^4-x^2+1# My fault not to remember the simple algebraic formula:
Find roots of x^6-×-1 - Socratic
2018年6月5日 · See roots below. x^6-x-1 =0 This degree 6 polynomial will have 6 real and/or complex roots. Using the ...
What is the integration of #(x^6+1)/(x+1)# - Socratic
2018年3月23日 · 722 views around the world You can reuse this answer ...
How do you solve #log (x+6)=1-log(x-5)#? - Socratic
2016年3月20日 · x=(-1+-sqrt(161))/2 1. Bring all the logs to the left side of the equation. log(x+6)=1-log(x-5) log(x+6)+log(x-5)=1 2. Use the log property, log_color(purple)b(color(red)m*color(blue)n)=log_color(purple)b(color(red)m)+log_color(purple)b(color(blue)n), to rewrite the left side of the equation. log((x+6)(x-5))=1 3. Use the log property, log_color(purple)b(color(purple)b^color(orange)x)=color ...