Question

lf $\alpha ,\beta ,\gamma$ are the roots of $x^3+p{x}^2+qx+r=0$, then the value of $(1+{\alpha }^2)(1+{\beta }^2)(1+{\gamma }^2)$, is:

• A. $(r+p{)}^2+(q+1{)}^2$
• B. $(r-p{)}^2+(q+1{)}^2$
• C. $(1+p{)}^2+(1+q{)}^2$
• D. $(r-p{)}^2+(r-q{)}^2$

Answer 1

Answer: (B)

$(r-p{)}^2+(q+1{)}^2$

As $\alpha ,\beta ,\gamma$ are roots of $x^3+p{x}^2+qx+r=0$
Replacing by $x\to x-i$, we get
${(x-i)}^3+p{(x-i)}^2+q(x-i)+r=0\Rightarrow x^3-i+px-p-2pxi+qx-qi+r=0\Rightarrow x^3+px+qx-2pix-i-p-qi+r=0$

And $(\alpha +i),(\beta +i),(\gamma +i)$ are roots of this equation
$s_3=-i-p-qi+r=(\alpha +i)(\beta +i)(\gamma +i)$
$\Rightarrow (r-p)-i(q+1)=(\alpha +i)(\beta +i)(\gamma +i)$

Taking moduls, we get
${(r-p)}^2+{(q+1)}^2={(\alpha +i)}^2{(\beta +i)}^2{(\gamma +i)}^2\Rightarrow {(r-p)}^2+{(q+1)}^2=(1+{\alpha }^2)(1+{\beta }^2)(1+{\gamma }^2)$