Solve : $x^6 - 12x^5 + ax^4 + bx^3 + cx^2 + dx + 64 =0$ has all positive roots then the values of $a , b , c$ and $d$ are

Question

My try : Tried to solve by relation between coefficient and roots of n- degree equations but unable to proceed because of the variables. But just found the interesting thing that the signs of $a , b , c$ and $d$ are as same as in relation.

Answer : $a= 60, b= -160 ,c= 240 , d= -192$

Answer 1

Using Vieta's formula and AM-GM: $$x_1+x_2+x_3+x_4+x_5+x_6=12;\\ x_1x_2x_3x_4x_5x_6=64;\\ x_1+x_2+x_3+x_4+x_5+x_6\ge 6\sqrt[6]{x_1x_2x_3x_4x_5x_6} \Rightarrow x_1x_2x_3x_4x_5x_6 \le 64,$$ the equality occurs for $x_1=x_2=\cdots= x_6=2$. So, the equation is: $(x-2)^6=0.$

Expand and find the coefficients.

Answer 2

@farruhota 's approach is correct, but instead of expanding $\left(x-2\right)^{6}$ ,what you can do is use a little bit of permutation and combination: here's how: let the roots be A,B,C,D,E,F. now AB+BC+CD......(roots taken 2 at a time)= $\frac{a}{1}$ thus, $\frac{\left(AB+BC+CD......\right)}{15}\ge\left[\left(ABCDEF\right)^{5}\right]^{\frac{1}{15}}$

How 15, well six roots taken two at a time, then total possibilities are 6C2 = 15 and each root appears 5 times, so $\left(ABCDEF\right)^{5}$

Now, plug in the value of ABCDEF=64 (product of roots) and when you solve it you'll get AB+BC+CD......=a $\ge$ 60, but since roots are equal, equality will follow so you get a=60