Approach : a^three - b^three = (a-b)(a^2+ab+b^2) resolution: x^3 - 1/eight utilizing approach (x - half)(x^2+x*1/2+1/4) now, each x-half =0 x=1/2 or x^2+x*1/2+1/four=0 4x^2+2x+1=zero (multiplying all sides via four) with the aid of using quadratic method x = -1/four+root.Three/4i , -1/4-root.3/4i. As a consequence; resolution set 1/2 of , -1/four+root.Three/4i , -1/4-root.3/4i
Answers & Comments
if x = 1/2, this becomes 0 so (x - 1/2) is a factor:
x^3 - 1/8 = x^2 (x - 1/2) + 1/2 x(x - 1/2) + 1/4 (x - 1/2)
= (x - 1/2)(x^2 + 1/2 x + 1/4)
The formula you need to use is a^3 - b^3 = (a-b)(a^2+ab+b^2)
In this case, a is x and b is 1/2
Therefore, x^3 - 1/8 = (x - 1/2)(x^2 + x/2 + 1/4)
Approach : a^three - b^three = (a-b)(a^2+ab+b^2) resolution: x^3 - 1/eight utilizing approach (x - half)(x^2+x*1/2+1/4) now, each x-half =0 x=1/2 or x^2+x*1/2+1/four=0 4x^2+2x+1=zero (multiplying all sides via four) with the aid of using quadratic method x = -1/four+root.Three/4i , -1/4-root.3/4i. As a consequence; resolution set 1/2 of , -1/four+root.Three/4i , -1/4-root.3/4i
(x – 1/2)(x² + x/2 + 1/4)
Expand it out and you have:
x³ + x²/2 + x/4 – x²/2 – x/4 – 1/8 = x³ – 1/8
there's kind of a trick tothese to make the middle terms drop out:
(x-1/2)(x^2 + x/2 + 1/4)
see how the 1/2*x^2 and the -1/2*x^2 make these terms drop out!
good luck!
formula : a^3 - b^3 = (a-b)(a^2+ab+b^2)
solution:
x^3 - 1/8
applying formula
(x - 1/2)(x^2+x*1/2+1/4)
now,
either x-1/2 =0
x=1/2
or
x^2+x*1/2+1/4=0
4x^2+2x+1=0 (multiplying both sides by 4)
by using quadratic formula
x = -1/4+root.3/4i , -1/4-root.3/4i.
hence; solution set {1/2 , -1/4+root.3/4i , -1/4-root.3/4i}