+0  
 
+2
74
3
avatar+290 

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
avatar+867 
+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
avatar
+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
avatar+290 
+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

13 Online Users

avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.