Question

Find the values of
(1) $x^3+x^2-x-22$ when $x=1+2i$
(2) $x^3-3x^2-8x+15$ when $x=3+i$
(3) $x^3-a{x}^2+2a^{2}x+4a^3$ when $\frac{x}{a}=1-\sqrt{-3}$

Answer 1

1) $x=1+2i$
$⟹x^3+x^2-x-22={(1+2i)}^3+{(1+2i)}^2-(1+2i)-22$
$=[1+8i^3+3\left(2i\right)+3{\left(2i\right)}^2]+[1+4i^2+4i]-1-2i-22$$=1+8i+6i-12+1-4+4i-1-2i-22$
[$\because i^2=-1$ and $i^3=-i$ ]
$=0i-37$
$=-37$
2) $x=3+i$
$⟹x^3-3x^2-8x+15={(3+i)}^3{-3(3+i)}^2-8(3+i)+15$
$=[27+i^3+3{\left(3\right)}^{2}i+3\times 3{\left(i\right)}^2]-3[9+i^2+6i]-24-8i+15$
$=[27-i+27i-9]-3[9-1+6i]-9-8i$
$=26i+18-24-18i-9-8i$
$=26i-26i+18-33$
$=-15$
3) $x=a(1-\sqrt{3}i)$
$x^3-a{x}^2+2a^{2}x+4a^3=a^3[{\left(\frac{x}{a}\right)}^3-{\left(\frac{x}{a}\right)}^2+2\frac{x}{a}+4]$
${=a}^3[{(1-\sqrt{3}i)}^3-{(1-\sqrt{3}i)}^2+2(1-\sqrt{3}i)+4]$
${=a}^3[1-3\sqrt{3}{i}^3-3\sqrt{3}i+3\left(3i^2\right)-1[1+3i^2-2\sqrt{3}i]+6-2\sqrt{3}i]$
${=a}^3[1+3\sqrt{3}i-3\sqrt{3}-9-1[1-3-2\sqrt{3}i]+6-2\sqrt{3}i]$
${=a}^3[-8+2+2\sqrt{3}i+6-2\sqrt{3}i]$
${=a}^3[-8+8]$
$=0$