Hi, tonio thanks for the tip. I'm aware of it, although I'm trying to solve it using the method that my textbook uses.this is an example of the question using the factorisation and squaring of both sides: seems pretty simple aint it? except i can't seem to work it out for the one im doing

**EXAMPLE QUESTION**
If $\displaystyle \alpha,\beta,\gamma,\delta$ are the roots of $\displaystyle x^{3}+2x^{2}-2x+3=0$, form the equation whose roots are:$\displaystyle \alpha^{2},\beta^{2},\gamma^{2},\delta^{2}$

$\displaystyle Let$

$\displaystyle y=x^{2}$

$\displaystyle x=y^\frac{1}{2}$

$\displaystyle (y^\frac{1}{2})^{3}+2(y^\frac{1}{2})^{2}-2(y^\frac{1}{2})+3=0$

$\displaystyle y^\frac{3}{2}+2y-2y^\frac{1}{2}+3=0$

$\displaystyle y^\frac{3}{2}-2y^\frac{1}{2}=-3-2y$

$\displaystyle y^\frac{1}{2}(y-2)=-3-2y$

square both sides to get

$\displaystyle y(y-2)^{2}=9+12y+4y^{2}$

$\displaystyle y^{3}-4y^{2}+4y=9+12y+4y^{2}$

$\displaystyle y^{3}-8y^{2}-8y-9=0$

I was wondering with my original question..would i be able to factorise out so that squaring both sides would eliminate a surd

Thanks for your help