System : a^3 - b^three = (a-b)(a^2+ab+b^2) resolution: x^3 - 1/8 making use of system (x - 1/2)(x^2+x*1/2+1/4) now, both x-1/2 =0 x=1/2 or x^2+x*1/2+1/four=0 4x^2+2x+1=zero (multiplying all sides through four) with the aid of utilizing quadratic method x = -1/4+root.3/4i , -1/four-root.3/4i. As a result; solution set half of , -1/four+root.Three/4i , -1/four-root.3/4i
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Verified answer
a^3 - b^3 = (a - b)(a^2 + a*b + b^2)
x^3 - 1/8 = (x)^3 - (1/2)^3
x^3 - 1/8 = (x - 1/2)(x^2 + x/2 + 1/4)
System : a^3 - b^three = (a-b)(a^2+ab+b^2) resolution: x^3 - 1/8 making use of system (x - 1/2)(x^2+x*1/2+1/4) now, both x-1/2 =0 x=1/2 or x^2+x*1/2+1/four=0 4x^2+2x+1=zero (multiplying all sides through four) with the aid of utilizing quadratic method x = -1/4+root.3/4i , -1/four-root.3/4i. As a result; solution set half of , -1/four+root.Three/4i , -1/four-root.3/4i
(2x-1)(x²+1/2x+1/4)
factorizing (x²+1/2x+1/4) will get you complex root.
In case you want the complex root...
-1/4 + 1/4√3 i
and
-1/4 - 1/4√3 i
a3 – b3 = (a – b)(a2 + ab + b2)
x^3 - (1/2)^3 = (x - 1/2)(x^2 +x/2 + 1/4)
(x minus 1/2) cubed