We have a fifth degree polynomial:
[tex]x^5-3x^4+mx^3+nx^2+px+q=0[/tex]This polinomial has 3 real roots, that can be expressed as: log2(a), log2(b) and log2(c).
Also, it has two imaginary roots, one of which is di (they have to be conjugate, so the other imginary root is -di).
We have to show that abc = 8.
If we consider the information given, we have some information about all the roots.
We can rewrite the polynomial in factorized form as:
[tex]\begin{gathered} (x-\log _2a)(x-\log _2b)(x-\log _2c)(x-di)(x+di)=0 \\ (x-\log _2a)(x-\log _2b)(x-\log _2c)(x^2+d^2)=0 \end{gathered}[/tex]As the polynomial is defined for real numbers, we can write a polynomial with only the real roots as:
[tex](x-\log _2a)(x-\log _2b)(x-\log _2c)=0[/tex]Then, we can relate the roots as:
[tex]\begin{gathered} 2^{(x-\log _2a)(x-\log _2b)(x-\log _2c)}=2^0 \\ 2^{(x-\log _2a)}\cdot2^{(x-\log _2b)}\cdot2^{(x-\log _2c)}=1 \\ \frac{2^x}{2^{\log_2a}}\cdot\frac{2^x}{2^{\log_2a}}\cdot\frac{2^x}{2^{\log_2a}}=1 \\ \frac{2^{3x}}{a\cdot b\cdot c}^{}=1 \\ abc=2^{3x} \\ abc=2^3\cdot2^x \\ abc=8\cdot2^x \end{gathered}[/tex]