Find all real numbers \(x\) such that \(\left(\dfrac{x}{3}\right)^3-3x^2+81x-729=25+2(5)(3)+9.\)

TheMathCoder
Apr 26, 2018

#1**+2 **

On the left side of the equation, we can tell that that we can easily factor it.

It is a perfect cube.

We use \((x-y)^3=x^3-3x^2y+3xy^2-y^3\)

Then we have:

\((\frac{x}{3}-3)^3=64\)

\(\frac{x}{3}-3=4 \)

x = 21

I hope this helped,

Gavin

GYanggg
Apr 27, 2018

#2**+1 **

**Gavin: Your solution dosen't balance the equation!!!!**

Solve for x:

x^3/27 - 3 x^2 + 81 x - 729 = 64

Subtract 64 from both sides:

x^3/27 - 3 x^2 + 81 x - 793 = 0

The left hand side factors into a product with three terms:

1/27 (x - 39) (x^2 - 42 x + 549) = 0

Multiply both sides by 27:

(x - 39) (x^2 - 42 x + 549) = 0

Split into two equations:

x - 39 = 0 or x^2 - 42 x + 549 = 0

Add 39 to both sides:

x = 39 or x^2 - 42 x + 549 = 0

Subtract 549 from both sides:

x = 39 or x^2 - 42 x = -549

Add 441 to both sides:

x = 39 or x^2 - 42 x + 441 = -108

Write the left hand side as a square:

x = 39 or (x - 21)^2 = -108

Take the square root of both sides:

x = 39 or x - 21 = 6 i sqrt(3) or x - 21 = -6 i sqrt(3)

Add 21 to both sides:

x = 39 or x = 21 + 6 i sqrt(3) or x - 21 = -6 i sqrt(3)

Add 21 to both sides:

**x = 39 or x = 21 + 6 i sqrt(3) or x = 21 - 6 i sqrt(3)**

Guest Apr 27, 2018

#3**+2 **

Thank you guys! GYanggg, even though you got it wrong, good job, I really appreciate your and guest’s help.

TheMathCoder
Apr 27, 2018