Question

1. $x^3 + 125 = 0$
2. $8x^5 + 5x^2 = 4x^2$
3. $x^3 - 216 = 0$
4. $16x^3 = 54$
5. $x^4 + x^2 - 42 = 0$
6. $x^4 + 3 = 13 - 9x^4$

Explanation:

Step1: Solve $x^3 + 125 = 0$

Rewrite as sum of cubes: $x^3 + 5^3 = 0$
Use $a^3+b^3=(a+b)(a^2-ab+b^2)$:
$$(x+5)(x^2-5x+25)=0$$
Set each factor to 0:

1. $x+5=0 \implies x=-5$
2. $x^2-5x+25=0$

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=-5,c=25$:
$$x=\frac{5\pm\sqrt{25-100}}{2}=\frac{5\pm\sqrt{-75}}{2}=\frac{5\pm5i\sqrt{3}}{2}$$

Step2: Solve $8x^5 + 5x^2 = 4x^2$

Rearrange to standard form:
$$8x^5 + x^2 = 0$$
Factor out $x^2$:
$$x^2(8x^3+1)=0$$
Rewrite $8x^3+1$ as sum of cubes: $x^2((2x)^3+1^3)=0$
Use $a^3+b^3=(a+b)(a^2-ab+b^2)$:
$$x^2(2x+1)(4x^2-2x+1)=0$$
Set each factor to 0:

1. $x^2=0 \implies x=0$ (double root)
2. $2x+1=0 \implies x=-\frac{1}{2}$
3. $4x^2-2x+1=0$

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=4,b=-2,c=1$:
$$x=\frac{2\pm\sqrt{4-16}}{8}=\frac{2\pm\sqrt{-12}}{8}=\frac{1\pm i\sqrt{3}}{4}$$

Step3: Solve $x^3 - 216 = 0$

Rewrite as difference of cubes: $x^3 - 6^3 = 0$
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$:
$$(x-6)(x^2+6x+36)=0$$
Set each factor to 0:

1. $x-6=0 \implies x=6$
2. $x^2+6x+36=0$

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=1,b=6,c=36$:
$$x=\frac{-6\pm\sqrt{36-144}}{2}=\frac{-6\pm\sqrt{-108}}{2}=-3\pm3i\sqrt{3}$$

Step4: Solve $16x^3 = 54$

Divide both sides by 2:
$$8x^3 = 27$$
Rearrange to difference of cubes: $(2x)^3 - 3^3 = 0$
Use $a^3-b^3=(a-b)(a^2+ab+b^2)$:
$$(2x-3)(4x^2+6x+9)=0$$
Set each factor to 0:

1. $2x-3=0 \implies x=\frac{3}{2}$
2. $4x^2+6x+9=0$

Use quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ where $a=4,b=6,c=9$:
$$x=\frac{-6\pm\sqrt{36-144}}{8}=\frac{-6\pm\sqrt{-108}}{8}=\frac{-3\pm3i\sqrt{3}}{4}$$

Step5: Solve $x^4 + x^2 - 42 = 0$

Let $u=x^2$, substitute:
$$u^2 + u - 42 = 0$$
Factor the quadratic:
$$(u+7)(u-6)=0$$
Solve for $u$:

1. $u+7=0 \implies u=-7$
2. $u-6=0 \implies u=6$

Substitute back $u=x^2$:

1. $x^2=-7 \implies x=\pm i\sqrt{7}$
2. $x^2=6 \implies x=\pm\sqrt{6}$

Step6: Solve $x^4 + 3 = 13 - 9x^4$

Rearrange to standard form:
$$10x^4 - 10 = 0$$
Divide by 10:
$$x^4 - 1 = 0$$
Factor as difference of squares: $(x^2-1)(x^2+1)=0$
Factor further:
$$(x-1)(x+1)(x^2+1)=0$$
Set each factor to 0:

1. $x-1=0 \implies x=1$
2. $x+1=0 \implies x=-1$
3. $x^2+1=0 \implies x=\pm i$

Answer:

1. $x=-5$, $x=\frac{5\pm5i\sqrt{3}}{2}$
2. $x=0$ (double root), $x=-\frac{1}{2}$, $x=\frac{1\pm i\sqrt{3}}{4}$
3. $x=6$, $x=-3\pm3i\sqrt{3}$
4. $x=\frac{3}{2}$, $x=\frac{-3\pm3i\sqrt{3}}{4}$
5. $x=\pm\sqrt{6}$, $x=\pm i\sqrt{7}$
6. $x=1$, $x=-1$, $x=\pm i$