Question

1. $sqrt{x^3} = sqrt3{-27}$
2. $sqrt{b^3} = sqrt3{512}$
3. $8t^2 + 6 = 70$
4. $256 = -\frac{1}{2}p^8$
5. $250 = 2n^5$
6. $6w^4 - 9 = 153$
7. the volume of a number cube is 0.343 cubic inches. what is the length of the number cube?

Explanation:

Step1: Solve $\sqrt[3]{x^3} = \sqrt[3]{-27}$

$x = \sqrt[3]{-27} = -3$

Step2: Solve $\sqrt[3]{b^3} = \sqrt[3]{512}$

$b = \sqrt[3]{512} = 8$

Step3: Isolate $8r^3$ in $8r^3+6=70$

$8r^3 = 70 - 6 = 64$

Step4: Solve for $r$

$r^3 = \frac{64}{8} = 8 \implies r = \sqrt[3]{8} = 2$

Step5: Isolate $-\frac{1}{2}p^4$ in $256 = -\frac{1}{2}p^4$

$p^4 = 256 \times (-2) = -512$ (No real solution; if considering complex: $p = \sqrt[4]{-512} = 4(-1)^{1/4}$)

Step6: Isolate $x^3$ in $250 = 2x^3$

$x^3 = \frac{250}{2} = 125$

Step7: Solve for $x$

$x = \sqrt[3]{125} = 5$

Step8: Isolate $6w^3$ in $6w^3-9=153$

$6w^3 = 153 + 9 = 162$

Step9: Solve for $w$

$w^3 = \frac{162}{6} = 27 \implies w = \sqrt[3]{27} = 3$

Step10: Solve for cube side length

Let $s$ = side length, $V = s^3 = 0.343$. $s = \sqrt[3]{0.343} = 0.7$

Answer:

1. $x = -3$
2. $b = 8$
3. $r = 2$
4. No real solution; complex: $p = 4(-1)^{1/4}$
5. $x = 5$
6. $w = 3$
7. $0.7$ inches