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# Binomial Theorem

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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
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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

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

Subtract 549 from both sides:

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

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)

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

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

Guest Apr 27, 2018
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Thank you guys! GYanggg, even though you got it wrong, good job, I really appreciate your and guest’s help.

TheMathCoder  Apr 27, 2018