Do you mean your teacher can't solve cubics either?
x<SUP>3</SUP>+x-2=0 so test divisors of 2, in particular, 1 works, so
x<SUP>3</SUP>+x-2=(x-1)(x<SUP>2</SUP>+x+2)=0 and the discriminant for the second factor is -7 so x=1 and the points are (1,1), (1,-1). If no divisor worked you could use...