From the given equation
#3x^(2/3)+x^(1/3)-2=0#
The solution goes like this
Let #w=x^(1/3)#
#3x^(2/3)+x^(1/3)-2=0#
we can write this equation this way
#3(x^(1/3))^2+(x^(1/3))^1-2=0#
also
#3(w)^2+1*(w)^1-2=0#
also let #a=3# and #b=1# and #c=-2#
#w=(-b+-sqrt(b^2-4ac))/(2a)#
#w=(-1+-sqrt(1^2-4*3*(-2)))/(2*3)#
#w=(-1+-sqrt(1+24))/6#
#w=(-1+-sqrt25)/6#
#w=(-1+-5)/6#
we have two values for w:
#w_1=(-1+5)/6=4/6=2/3# and #w_2=(-1-5)/6=-6/6=-1#
But #w=x^(1/3)#
and #x=w^3#
Using #w=2/3#
#x=w^3=(2/3)^3=8/27#
#x=8/27#
Using #w=-1#
#x=w^3=(-1)^3=-1#
#x=-1#
check at #x=8/27#
#3x^(2/3)+x^(1/3)-2=0#
#3(8/27)^(2/3)+(8/27)^(1/3)-2=0#
#3(4/9)+2/3-2=0#
#4/3+2/3-2=0#
#6/3-2=0#
#2-2=0#
#0=0#
#x=8/27# is a root
check at #x=-1#
#3x^(2/3)+x^(1/3)-2=0#
#3(-1)^(2/3)+(-1)^(1/3)-2=0#
#3(1)-1-2=0#
#3-3=0#
#0=0#
#x=-1# is a root
God bless....I hope the explanation is useful.
