Question 17951
8x^6+7x^3-1=0

Put y=x^3

Then we have

8y^2+7y-1=0

8y^2+8y-1y-1=0

8y(y+1)-1(y+1)=0

(8y-1)(y+1)=0

y=1/8 or -1

Therefore,x^3=1/8 or -1

x=1/2 or [cube root of -1]

x=1/2 or [1,{{{(1 +- sqrt(3)i)/2}}}]

<br>Thus we get 4 roots for the equation,
{{{ x=(1/2)}}}
{{{ x = (-1)}}}
{{{ x = (1 +- sqrt(3)i)/2 }}}
<P>
Hope this helps,
Prabhat