Question 4625
{{{x^6+7x^3-8=0}}}, think of this as a quadratic {{{y^2+7y-8=0}}}, where{{{y=x^3}}}.


so, factorise to give (y+8)(y-1)=0, so then we say that either y+8=0 OR y-1=0.


Hence y=-8 or y=+1


So, {{{x^3=-8}}} OR {{{x^3=1}}}, which means that the possible answers are:


x=-2 or x=+1


jon