Question

1. solve $27x^3 - 343 = 0$.

$x = -3$ and $x = \frac{-7 \pm 7i\sqrt{3}}{6}$
$x = \frac{7}{3}$ and $x = \frac{-7 \pm 7i\sqrt{3}}{6}$
$x = 3$ and $x = \frac{-1 \pm 5i\sqrt{3}}{6}$
$x = -7$ and $x = -7 \pm 7i\sqrt{3}$

Explanation:

Step 1: Rewrite the equation

We have the equation \( 27x^3 - 343 = 0 \). We can rewrite this as \( (3x)^3 - 7^3 = 0 \).

Step 2: Use the difference of cubes formula

The difference of cubes formula is \( a^3 - b^3=(a - b)(a^2+ab + b^2) \). Here, \( a = 3x \) and \( b = 7 \), so we get:
\( (3x - 7)((3x)^2+(3x)(7)+7^2)=0 \)
\( (3x - 7)(9x^2 + 21x + 49)=0 \)

Step 3: Solve for \( x \) from each factor

• For the first factor \( 3x - 7 = 0 \), we solve for \( x \):

\( 3x=7 \)
\( x=\frac{7}{3} \)

• For the second factor \( 9x^2 + 21x + 49 = 0 \), we use the quadratic formula \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \), where \( a = 9 \), \( b = 21 \), and \( c = 49 \).

First, calculate the discriminant \( \Delta=b^2-4ac=(21)^2-4\times9\times49=441 - 1764=- 1323 \)
Then, \( x=\frac{-21\pm\sqrt{-1323}}{2\times9}=\frac{-21\pm\sqrt{1323}i}{18} \)
Simplify \( \sqrt{1323}=\sqrt{441\times3}=21\sqrt{3} \), so:
\( x=\frac{-21\pm21\sqrt{3}i}{18}=\frac{-7\pm7\sqrt{3}i}{6} \)

Answer:

\( x = \frac{7}{3} \) and \( x = \frac{-7 \pm 7i\sqrt{3}}{6} \)