+295 Find all real numbers \(x\) such that \(\left(\dfrac{x}{3}\right)^3-3x^2+81x-729=25+2(5)(3)+9.\)
+983 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
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)
+295 Thank you guys! GYanggg, even though you got it wrong, good job, I really appreciate your and guest’s help.
