Note: As you may have unintentionally missed to add the value choices. But, I would make sure to explain the concept so that you may improve your understanding in terms of solving these type of questions.
Answer:
Any value other than the values [tex]x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}[/tex] will not be a solution of [tex]8x^3\:-\:1\:=\:0[/tex].
Step-by-step explanation:
Considering the equation
[tex]8x^3\:-\:1\:=\:0[/tex]
Steps to solve the equation
[tex]8x^3-1=0[/tex]
[tex]\mathrm{Add\:}1\mathrm{\:to\:both\:sides}[/tex]
[tex]8x^3-1+1=0+1[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]x^3=\frac{1}{8}[/tex]
[tex]\mathrm{Divide\:both\:sides\:by\:}8[/tex]
[tex]\frac{8x^3}{8}=\frac{1}{8}[/tex]
[tex]\mathrm{Simplify}[/tex]
[tex]x^3=\frac{1}{8}[/tex]
As
[tex]\mathrm{For\:}x^3=f\left(a\right)\mathrm{\:the\:solutions\:are\:}x=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}[/tex]
[tex]x=\sqrt[3]{\frac{1}{8}},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1+\sqrt{3}i}{2},\:x=\sqrt[3]{\frac{1}{8}}\frac{-1-\sqrt{3}i}{2}[/tex]
So,
[tex]x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}[/tex]
Therefore,
Any value other than the values [tex]x=\frac{1}{2},\:x=-\frac{1}{4}+i\frac{\sqrt{3}}{4},\:x=-\frac{1}{4}-i\frac{\sqrt{3}}{4}[/tex] will not be a solution of [tex]8x^3\:-\:1\:=\:0[/tex].
Keywords: solution, value
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