If $\tan A, \tan B, \tan C$ are roots of $x^3-ax^2+b=0$, find $(1+\tan^2 A)(1+\tan^2B)(1+\tan^2C)$ Here are all the results I got
$$\tan(A+B+C)=a-b$$
And
$$(1+\tan^2A)(1+\tan^2B)(1+\tan^2C)=(\frac{1}{\cos A\cos B\cos C})^2$$
And
$$\cot A+\cot B + \cot C=0$$
How should I use these results?
 A: It is a trick. Let $p(x) =  x^3-ax^2+b$, then   $$p(x)= (x-\tan A)(x-\tan B)(x-\tan C)$$
Now you need $$p(i)\cdot p(-i)$$
A: Write $x=\tan A, y=\tan B, z=\tan C$.
By Vieta’s formula, we have
$$x+y+z=a,\ \ xy+yz+zx=0,\ \ xyz=-b.$$
Now you want to find the value of $$(1+x^2)(1+y^2)(1+z^2)=1+x^2+y^2+z^2+x^2y^2+y^2z^2+z^2x^2+x^2y^2z^2.$$
Note that
$$a^2=(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx),$$
so $x^2+y^2+z^2=a^2$. Similarly,
$$0=(xy+yz+zx)^2=x^2y^2+y^2z^2+z^2x^2+2xyz(x+y+z)$$
implies $x^2y^2+y^2z^2+z^2x^2=2ab$. Hence
$$(1+x^2)(1+y^2)(1+z^2)=1+a^2+2ab+b^2=1+(a+b)^2.$$
A: So since others have shown workings I will do mine - Aqua's answer is equivalent and Toby Mak has a different way of doing the same thing.
Let $y=x^2+1$ so that $x^2=y-1$ and substitute as much as possible in the original cubic: $$x(y-1)-a(y-1)+b=0$$
Rearrange to isolate $x$ and obtain: $$x(y-1)=a(y-1)+b=ay-(a+b)$$
Now square this:
$$x^2(y-1)^2=(y-1)^3=\left(ay-(a+b)\right)^2$$ so that $$y^3-3y^2+3y+1-a^2y^2+2a(a+b)y-(a+b)^2=0$$
Now for the product of the roots you need just the constant term, and since the coefficient of $y^3$ is $+1$ you need the negative of the constant term, hence $(a+b)^2-1$
This uses only manipulation of the original equation and Vieta's relation in the final cubic.
This method, where applicable is completely algebraic, avoids working in any algebraic or analytic context beyond the world of the coefficients of the cubic, and avoids explicit manipulation of symmetric polynomials in the workings. I have always found these things an advantage. It uses no particular properties of the tan function, but I couldn't see how that would make a simpler solution. It only uses that the roots of two equations can be related.
A: Set $x = 1 + y^2 \Rightarrow y = ±\sqrt{x - 1}$ so that $1 + \tan^2 A, 1 + \tan^2 B, 1 + \tan^2 C$ give the roots of $x^3 - ax^2 + b = 0$.
This gives $±(x-1)^{3/2} - a(x-1) +b=0$. By Vieta, the product of the roots is the constant term, but we do not have a polynomial yet. Squaring both sides of $±(x  - 1)^{3/2} = a(x-1) - b$, the constant term is $(-1)^3 - (-a-b)^2 = -1 -(a+b)^2$ if we take the positive branch, and $(-a-b)^2 - (-1) = 1 + (a+b)^2$ if we take the negative branch. However, since $(1 + \tan^2 A)(1 + \tan^2 B)(1 + \tan^2 C)$ is always positive, the answer is $1 + (a+b)^2$.
A: It is basically an algebraic problem:
If $p,q,r$ are the roots of $$x^3-ax^2+b=0$$
it is sufficient to find the cubic equation whose roots are $w=p^2+1,y=q^2+1,z=r^2+1$
From $ap^2-b=p^3,$ we have
$$(ap^2-b)^2=(p^3)^2=(p^2)^3$$
Replace $p^2$ with $w-1$  to find $$(a(w-1)-b)^2=(w-1)^3$$
$$\iff w^3+(\cdots)w^2+(\cdots)w-1-(a+b)^2=0$$
Obviously,  we shall reach at the same equation if start with $y=q^2+1$ or $z=r^2+1$
Using Vieta's  formula, $$wyz=-\dfrac{-1-(a+b)^2}1$$
See also :

*

*Finding $\sum_{k=0}^{n-1}\frac{\alpha_k}{2-\alpha_k}$, where $\alpha_k$ are the $n$-th roots of unity


*If $\alpha$, $\beta$, $\gamma$ and $\delta$ be the roots of the polynomial $x^4+px^3+qx^2+rx+s$, prove the following.
