Question

If $a,b,c$ are the roots of $x^3+8=0$, then the equation whose roots are $a^2,b^2,c^2$ is-

• A. $x^3-8=0$
• B. $x^3+64=0$
• C. $x^3+16=0$
• D. $x^3-64=0$

Answer 1

The correct option is D $x^3-64=0$

$x^3+8=0rootsarea,b,ca+b+c=0ab+bc+ca=0abc=-8newrootsare{a}^2,b^2,c^2{a}^2+b^2+c^2={(a+b+c)}^2-2(ab+bc+ac)=0a^2{b}^2+b^2{c}^2+a^2{c}^2={(ab+bc+ac)}^2-2({ab}^{2}c+ba{c}^2+a^{2}bc)=0-2\left(abc\right)(a+b+c)=0a^2{b}^2{c}^2={\left(abc\right)}^2=64\therefore equationis{x}^3+0x^2+0x-64=0\therefore x^3-64=0$