Question

The given equation is ( x^{2 / 3}+x^{1 / 3}-2=0 )
Put ( x^{1 / 3}=y, ) then ( y^{2}+y-2=0 )
( Rightarrow(y-1)(y+2)=0 )
( Rightarrow y=1 ) or ( y=-2 )
( Rightarrow x^{1 / 3}=1 quad ) or ( quad x^{1 / 3}=-2 )
( therefore x=(1)^{3} ) or ( x=(-2)^{3}=-8 )
Hence, the real roots of the given equations are 1,-8

# The real roots of the equation r2/3+x/3 - 2 = 0) are (a) 1,8 (b) -1,-8 (c) 1,8 (d) 1,-8

Solution